PLEASE HELPPP. I'VE TRYING THIS QUESTION FOR 2 HRS & THE SOLUTIONS WERE WRONG! Reddit r/askmath

Рет қаралды 93,139

6 ай бұрын

We will solve a hard trigonometric equation 2sin(x)-2sqrt(3)cos(x)+sqrt(3)tan(x)-3=0. This equation is hard because it requires a very tricky factoring step. You will usually learn this kind of equation in a trigonometry or a precalculus class. Subscribe to ‪@bprpmathbasics‬ for more essential math tutorials.
I saw this question from Reddit r/askmath: / lafiyl1pol
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#math #algebra #mathbasics

Пікірлер: 141

@bprpmathbasics 6 ай бұрын

Try sin(πx)=sin(x) Answer: kzfaq.info/get/bejne/Y8eVl6tqtdvXomg.htmlsi=LXV32GBtrQIS8lFz

@gg-96 6 ай бұрын

x = 0

@lexyeevee 6 ай бұрын

π = 1

@atharvabhardwaj3830 6 ай бұрын

@junkgum 6 ай бұрын

Someone's wrong by 316.17..% or off by 2.1617..

@user-lj6mp8kb9i 6 ай бұрын

General soln is nπ/(π-(-1)ⁿ)

@Ruija27 6 ай бұрын

Before watching, I wonder if it's possible to twist this into something that fits the quadratic formula

@BigDBrian 6 ай бұрын

Another commenter pointed out that substituting t = tan (x/2) turns the problem into a polynomial

@septuamticsaneeh8891 6 ай бұрын

@@BigDBriando we have to apply double angle formula to both sin x and cos x in that case?

@chengkaigoh5101 6 ай бұрын

@@BigDBrianweisstrass substitution 🙏

@ngc-fo5te 6 ай бұрын

@@chengkaigoh5101No such thing.

@hake4972 6 ай бұрын

You could simply write tanx as sinx/cosx and multiply the equation by cosx. Also cos^2x equals to 1- sin^2x. Assume sinx as u. U get a quadratic in u.

@holyshit922 6 ай бұрын

For those who like polynomial equations t = tan(x/2) substitution and for fans of Euler's substitutions cos(x) = (1-sin(x))u

@EyeSooGuy 6 ай бұрын

Hey is Euler pronounced as “oy-ler”? Just wondering.

@holyshit922 6 ай бұрын

@@EyeSooGuy Yes, because he was from german language part of Switzerland What language is spoken in Basel ?

@EyeSooGuy 6 ай бұрын

@@l-_-lShadowCat aha. What I thought. Thanks.

@picup30296 6 ай бұрын

I tried the t-formula. It is tedious to work out the factorization, but it is so satisfying. Also notes for the pi/6 answers, you would get t=+-✓7+4✓3, which is 5π÷12.

@brian554xx 6 ай бұрын

I like to think of them as 2, 5, 7, and 8 times pi/6

@moxbroker 6 ай бұрын

So much trig flowed back into my brain from this one question

@shadowflamelightburst4503 6 ай бұрын

I worked it out before watching the video and was terrified when you pulled out the square root of three divided by cos(x) and then everything was resolved when you finally got it down to both equations. I originally just factored in my head to get to: (2cos(x)+sq3)(tan(x)-sq3)=0

@soumilbinhani8803 6 ай бұрын

If you see the only multiples where the √3 of cos and tan can be eliminated are multiples of 30° or π/6 and if you see coefficients we need cos to be √3/2 or -√3/2 that is only way to eliminate the √3 coefficient of cos so just comparing that answer we can se it needs to be -√3/2 to adjust the -3 at the end so it has to be x= 5π/6

@JubeiKibagamiFez 6 ай бұрын

I'm happy I found this channel, but all these maths are way out of my league all the way in another universe.

@johndickinson82 6 ай бұрын

You’ll get there with practice

@DanoshTech 6 ай бұрын

@@johndickinson82and alot of formulas 😅

@tomekk.1889 6 ай бұрын

I assume you're young. You'll be learning this in high school it's not as complicated as it looks

@JubeiKibagamiFez 6 ай бұрын

@@tomekk.1889 No. Old and working full time.

@sameepsingh4095 6 ай бұрын

Ah this was a easy one You will get used to these by the end of your 10th grade ig

@Mr.Ali_Ryuji 6 ай бұрын

I remember to know which quadrant is positive or negative I use abbreviation of ASTC. Add Sugar To Coffee. First quadrant all positive, second quadrant only Sin positive, third quadrant only Tan positive and last quadrant only Cos positive.

@stephenbeck7222 6 ай бұрын

All Students Take Calculus - good propaganda for students at this level to take more maths.

@nandkumars.vanamali1336 6 ай бұрын

If u go clock-wise Its just ACTS starting from first quadrant,4,3 to 2nd

@adixxx3056 4 ай бұрын

We called them CAST diagrams where you start in the bottom right and then label the quadrants anticlockwise

@EyeSooGuy 6 ай бұрын

Looks like analytic trigonometry! Wow.

@kylecow1930 6 ай бұрын

write a=sinx, b=cosx, then we have since by assumption b is not zero we can multiply through by it giving 2ab-2r3b^2+r3a-3b=0 if we collect the a terms we get a(2b+r3)-br3(2b+r3)=0 so we can factorise as (2b+r3)(a-br3)=0 which is now trivial to solve

@TranquilSeaOfMath 6 ай бұрын

Tore it up like the pro that you are! A nice exercise.

@bobbob-gg4eo 6 ай бұрын

Nice work! That one was tricky

@bprpmathbasics 6 ай бұрын

Thanks!

@pulkitsaggu 6 ай бұрын

i can not beleive on my self i did this ques on my own for first try man damn i have done a hard trignometry ques on my own for first time and this is a another level feeling

@ishaankumar4587 6 ай бұрын

This is not a hard question...try books like cengage trigonometry or something this question is school level at best

@Aditya-rt3zl 6 ай бұрын

@@ishaankumar4587JEE aspirants can solve this question in like 1 min😂

@pulkitsaggu 6 ай бұрын

Can anyone guess my age or class

@Nxck2440 6 ай бұрын

I couldn't solve it initially but when I clicked on the video to read the description I used the clue of factoring and then I got it :)

@codexcursors 6 ай бұрын

Algebra manipulation is my forte, and I was able to get down to the 2nd last step. Then when I saw how he got the answers by using those reference triangles, I got overwhelmed. Trigonometry sure is scary for algebra folks like me 😅

@NicolasMuñoz-p8g 6 ай бұрын

Same jaja. I had never seen such way of solving trigonometric equations

@sanskargupta7125 5 ай бұрын

just divide it by cosx and take common (2+root3 secx )(tanx-root3)=0

@NekoChan_TV 6 ай бұрын

Personnally I fatorised everybody by 4, and I got sin(x)/2 - sqrt(3)cos(x)/2 + sqrt(3)tan(x)/4 -3/4 = 0, note that 1/2 = sin (π/6+2kπ) or sin(π-π/6+2kπ), sqrt(3)/2 = cos(π/6+2kπ) or cos(-π/6+2kπ) and sqrt(3)=tan(π/6+2kπ), so it's get trivial that π/6+2kπ is solution, but way harder to prove if there's others solutions (or not) Now I'll watch the video xD

@bradleydillman8244 6 ай бұрын

(2+(square root(3)/cos x)) you can rewrite with this (2+square root (3) sec x) make it a clean. I don't like a fraction expression.

@juancarlossanchezveana1812 6 ай бұрын

Amazing

@lhopital2132 6 ай бұрын

Good evening Sir I have a doubt regarding modulus of x. That is can I use this identity for |x| = x sgn(x)

@Samir-zb3xk 6 ай бұрын

That is identity is true for all real numbers. |x|=x for positive x, |x|=-x for negative x, and zero doesn’t matter sgn(x)=1 for positive x, so xsgn(x)=x=|x| for positive x sgn(x)=-1 for negative x so xsgn(x)=-x=|x| for negative x And i think the convention is to say sgn(0)=0 but it might different elsewhere

@ThAlEdison 6 ай бұрын

I factored by writing sinx as tanx*cosx 2tanx*cosx-2sqrt(3)cosx+sqrt(3)tanx-sqrt(3)*sqrt(3)=0 is pretty easy to factor.

@no1brittbarbiefan 6 ай бұрын

Trains rights sisters! :transheart:

@barteqw 6 ай бұрын

Founded basic equations quite fast. I substituted tan(x)=sin(x)/cos(x), multiplying by cos(x) and factorized to form (sin(x) - sqrt(3)*cos(x))(2*cos(x)+sqrt(3))=0. Not very hard equation i think.

@kevinmadden1645 6 ай бұрын

You are absolutely correct!

@emilyesnyman 6 ай бұрын

I did something very similar

@Sam_73378 6 ай бұрын

I solved it by taking 4 common from the first 2 terms then writing it as 4sin(π/3-x) and doing the same with the last two by taking √3 common and converting tanx into sinx/cosx. From there i got two equations sin(π/3-x) =0 and cosx=0 and further solved it to get same answers

@ttkok11 6 ай бұрын

Engineer: Assume small angles: Sinx=tanx=x Cosx=1 2x-2sqrt3+sqrt3x-3=o (2+sqrt3)x =2sqrt3+3 x=6.4/3.7=1.6 or smth

@ohwow512 6 ай бұрын

This dude is in no way an engineer. I mean, he didn't even write 6,4/3.7 as "basically just 6/4"!

@ttkok11 6 ай бұрын

@@ohwow512 hey man even engineers have standards

@almightyhydra 6 ай бұрын

Those reference triangles are somewhat black magic. Is there a way to derive them from first principles?

@rantingrodent416 6 ай бұрын

Aren't they derived directly from the definitions of cos, sin and tan? The reference triangle is just "construct a unit triangle assuming cos _ and sin _"

@isaacormesher2851 6 ай бұрын

The 1/2/root3 one is from splitting an equilateral triangle of side length 2 in half, then the 1/1/root2 one is chopping a unit square in half along the diagonal.

@stephenbeck7222 6 ай бұрын

What is ‘first principles’? All math is from first principles. It’s just a matter of how far back to elementary school math you want him to do in one video. I think most people just memorize the solutions in the first quadrant for these basic values and remember which other quadrants have the extra solution depending on what trig function you have. Drawing a triangle in appropriate triangle is more fundamental than that but takes longer if you have to do it every time.

@stvp68 6 ай бұрын

Fellow college professor envious of a whiteboard eraser that actually works 😅

@maximocouretot00001 6 ай бұрын

very nice problem!

@SergeySvotin 5 ай бұрын

0:17 nope, thi is primitive one, you'd better factor 2cosx from first 2 members and sqrt3 from last 2, will be faster

@jasondominick5230 5 ай бұрын

I got the same answer with a less obvious method. 2sin(x) - 2sqrt(3)cos(x) + sqrt(3)tan(x) - 3 = 0 4(1/2)sin(x) - 4(sqrt(3)/2)cos(x) + 3(sqrt(3)/3)tan(x) - 3 = 0 4[sin(pi/6)sin(x) - cos(pi/6)cos(x)] + 3[tan(pi/6)tan(x) - 1] = 0 Using the sum of angles identity for cosine, -4cos(x + pi/6) - 3(2/sqrt(3))cos(x + pi/6)sec(x) = 0 cos(x + pi/6)[-4 - 2sqrt(3)sec(x)] = 0 Setting the factors equal to zero yields the same equations as in the video but in different forms. :)

@lhopital2132 6 ай бұрын

A problem regarding modulus of x. Problem is integral -2 to 2 |cosx| dx. Sir can you solve this problem by using above formula for modulus x. With out drawing graph for |cosx|

@Samir-zb3xk 6 ай бұрын

Definite integrals involving absolute modulus should be split into seperate integrals depending on where the function inside the absolute modulus is positive or negative It can be difficult to find full indefinite antiderivatives for absolute modulus functions because doing things like introducing the sign function, then integrating may cause continuity issues

@Samir-zb3xk 6 ай бұрын

Btw if you say ∫|cos(x)|dx=∫sgn(cos(x))cos(x)dx then do integration by parts you will get sgn(cos(x))sin(x)+c sgn(cos(x))sin(x) is a valid antiderivative of |cos(x)| because d/dx(sgn(cos(x))sin(x))=d/dx(abs(cos(x))tan(x))=|cos(x)| However, sgn(cos(x))sin(x) is not continuous, hence you cannot use the fundamental theorem of calculus on it to solve definite integrals.

@lhopital2132 6 ай бұрын

@@Samir-zb3xk but ∫ |x|/x dx = ∫ sgn(x) dx = |x| +c

@Samir-zb3xk 6 ай бұрын

@@lhopital2132 yea it works out nicely in that case, it does not work out nicely when you integrate |cos(x)|

@lhopital2132 6 ай бұрын

@@Samir-zb3xk OK @Samir-zb3xk thank you so much

@arjungupta8299 6 ай бұрын

Easy Problem Take 2cosx common from first 2 terms and root 3 from next 2 terms Factors will be ready, get the answer

@pauselab5569 6 ай бұрын

Never try to solve transcendental equations. Other than that, just sub sin(x) = a, cos(x)=b and start solving the equation. No reason you can’t solve an algebraic equation. Just not fun to deal with trig.

@togribble 6 ай бұрын

multiply everything by cos(x) and factor the -sqrt(3) as gcf from the -2sqrt(3)cos^2(x)-3cos(x) group. A bit easier to see this factor (-sqrt(3)) than the one you use. Answers must be checked due to possibility of cos(x)=0. ??

@ishaankumar4587 6 ай бұрын

Why would there be a possibilty for cosx=0

@vincenthills5024 6 ай бұрын

@@ishaankumar4587because you multiplied by cos x

@lokeshpatel6540 6 ай бұрын

Instead of writing tan x as sin x / cos x, I did it by writing sin x as tan x * cos x.

@tobybartels8426 6 ай бұрын

You don't have to rearrange the terms though. You can group them as they are, factor 2 out of the first group and factor √3/cos(x) out of the second group.

@firstname4337 6 ай бұрын

of course you don't have to rearrange them -- but it helps when you're teaching so the student can better see what's going on

@tobybartels8426 6 ай бұрын

@@firstname4337 : If you factor with the groups that bprp did, then of course you want to rearrange the terms to make that easier to follow. But if you factor with the groups in the order they came in, then it would only make it more confusing to rearrange them.

@closer_to_the_unknown 6 ай бұрын

It can be solved a little bit easier 2sin(x) - 2sqrt(3)cos(x) + sqrt(3)tan(x) - 3 = 0 2cos(x)(sin(x)/cos(x) - sqrt(3)) + sqrt(3)(tan(x) - 3/sqrt(3)) = 0 2cos(x)(tan(x) - sqrt(3)) + sqrt(3)(tan(x) - sqrt(3)) = 0 (tan(x) - sqrt(3))(2cos(x) + sqrt(3)) = 0 tan(x) = sqrt(3) or cos(x) = -sqrt(3)/2 x = pi/3 + pi*k x = +-5pi/6 + 2pi*k, k is whole number

@sebaberrios3071 5 ай бұрын

Hey I do not understand where appeared π/6? Can someone explain?

@davidoneil9148 5 ай бұрын

Well he used the special triange, 30-60-90, and the angle at which he put pi/6 is the 30degreee angle. Knowing your triangles is quite helpful. You can also find pi/6 from 30 degrees by doing 30 x pi/180 30pi/180 30 goes into 180 6 times pi/6 is ur answer

@hiteshpareeks 6 ай бұрын

i dont understand whats wrong with my factorisation can you please check x = pie/6 +n*pie,5*pie/6 + n*pie for n = integer

@okicek3016 6 ай бұрын

It should be +2pi×n, n∈Z (n is an integer) if im not mistaken. For example if you have sin(x+pi) it will give you -sin(x), while sin(x+2pi)=sin(x). However if this equation only had tangents it would've worked since tan(x+pi)=tan(x)

@eboone 6 ай бұрын

pie 🥧

@hiteshpareeks 6 ай бұрын

@@okicek3016 getting algebrically still same?….

@waffleonquaffle 6 ай бұрын

sin/cos is 2n*pi, tg/ctg is n*pi. Your second solution is incorrect, as it should be 5/6pi + 2n*pi OR 7/6pi + 2n*pi. Or if you wanna be fancy you can write it down as pi +- 1/6pi + 2n*pi

@lorenwilson8128 6 ай бұрын

Excel, plot the function, use solver to get the roots.

@AFSMG 6 ай бұрын

Brillante

@soshakobyan3123 6 ай бұрын

I solved this problem by completely different way. 😊

@durvius2657 6 ай бұрын

This was truly awesome, I pogged at the factoring of minus sqrt 3 cosine x part lol. Opened my eyes to the power of factoring like never before, thank you. :)

@zaidi3 6 ай бұрын

Assalamualaikum

@ddplayz9354 5 ай бұрын

We can break tan x and the answer will come upon factorisation of the formed equation... Right? I got π/3 and 5π/6 as answers.

@GFlCh 5 ай бұрын

So, the solutions are: 1) 5π/6 2) 7π/6 3) π/3 4) 4π/3 Why is only the first solution measured to the near side of the triangle, and the other 3 solutions are measured to the far side of the triangle?

@sovietcomrade7733 6 ай бұрын

I've trying lol

@leeshaocheng239 6 ай бұрын

b pen r pen is always smart and super

@MikehMike01 6 ай бұрын

I had a math professor who would say ‘you’ll never get abs from doing jumping jacks, no matter how much time you spend on it’ He was talking about the people spending 2 hours doing the wrong thing

@mahaveerchoudhury1445 4 ай бұрын

HOW Abt this approach 2(sinx-√3cosx) -3 = -√3tanx 2[-2,2] -3 = -√3tanx [-4,4] -3 = -√3tanx [-7,1] = -√3tanx [-1/√3 , 7/√3] = tanx Kindly tell if this approach is correct

@abhirupkundu2778 6 ай бұрын

How did the redditor not solve this💀💀

@ohwow512 6 ай бұрын

He either a, did not have access to a calculator to calculate inverse trig, or b, haven't learned about inverse trig yet

@WreckingYT 6 ай бұрын

how do you spend 2 hrs on that question

@looooonooooooooooooooooooooong 5 ай бұрын

slow ass brain maybe

@brandonk9299 6 ай бұрын

There is a much simpler approach: For simplicity lets represent tan(x) as t and cos(x) as c. Since we know that all are in terms of x, we can take advantage of the fact that sin(x) = tan(x)*cos(x), or sin (x)= c*t. I'll also use V3 to represent sqrt(3) for typing ease. Using that we have: 2 t c + V3 t - 2 V3 c - 3 = 0. Grouping: V3 t - 3 + 2 t c - 2 V3 c = (V3 t - 3) + (t - V3)*2 c = (t - V3)*V3 +(t - V3)*2 c = (t - V3)*(V3 + 2c) = 0 So we have t = tan (x) = V3 or sqrt(3) and c = cos (x) = - sqrt(3)/2

@thekingofgaming7497 6 ай бұрын

This question can be solved way faster by just staring at the problem for a bit and logicing your way through

@hanqnero 6 ай бұрын

Nice equation. We have one of such or similar type in senior high school math exam in Russia. Easy exam points for a lot of students.

@druhindatta1976 6 ай бұрын

Duh I checed it in one second with x = pi/3 and other multiples within

@84com83 6 ай бұрын

When are these, in my opinion complex, examples used in "real life"? Isn´t it just a method of how to use a non-English language?

@DravenFNM 6 ай бұрын

theyre extremely important in astronomy, navigation, and periodic functions used in physics, signal processing and quantum mechanics

@Retarc-me7ez 6 ай бұрын

Take 2 cos x out from the first 2 terms and √3 from last two terms. It'll be easier

@firstname4337 6 ай бұрын

WRONG

@arjungupta8299 6 ай бұрын

@@firstname4337HOW sir?

@wearron 6 ай бұрын

hellow

@user-ss6yz8vl4k 6 ай бұрын

You are not pnly agrait teacher but also respect students of any level. And guess what, there is always some thing new, at any level

@cosmos269 6 ай бұрын

I didn't understand how you are getting second graph for cosine? it should be on 4th quadrant?

@alibebek7967 6 ай бұрын

We are trying to find negative sqrt 3 over 2 not positive. On the 4th quadrant cosine is positive.

@Samir-zb3xk 6 ай бұрын

Cosine is negative in 2nd and 3rd quadrant

@benx2230 6 ай бұрын

I suspect the root of the problem is the questioners math skills are as poor as his English skills.

@heyfarhanirfan 6 ай бұрын

Yoooooooooooooooooooo…..

@PowerShellWizard 6 ай бұрын

There is a logical error in this solution. You can only divide by cos x if and only if you can guarantee that cos x is not zero, which the problem does not state. As such you are ruling out the possibility that cos x = 0 thus implying that x = pi/2 is not an acceptable solution. I know this sounds mute but division by zero is never acceptable and this should be explicitly stated

@bprpmathbasics 6 ай бұрын

I see what u mean. However I didn’t divide by cos(x), I factored out cos(x).

@PowerShellWizard 6 ай бұрын

@@bprpmathbasicsin stand corrected. However still at 0:52 when you expressed tanx = sinx/cosx you should have mentioned that this means x cannot be pi/2 :)

@lerarosalene 6 ай бұрын

@@PowerShellWizard pi/2 is incorrect not because of substitution tanx = sinx/cosx but because tan(pi/2) isn't defined at all.

@phiefer3 6 ай бұрын

@@PowerShellWizard There's no need to specify this though. The presence of tanx in the equation (and the fact that the equation is actually equal to something) means that x can only take on values in the domain of tan, which does not include pi/2 or 3pi/2. Which means that cosx cannot be 0.

@dQ__dU_dW 6 ай бұрын

When we write tan(x), its defined that at whatever value cos(x) = 0, should be excluded from the domain of tan(x) Even if its not mention we assume it to be otherwise the function isnt defined, its not a mistake more like a definiton

@wilhelmmeyer89 6 ай бұрын

The poor students! Many students just want to walk away when they here somebody say sin, cos or Pythagoras or something else related to more math than + and -. Pythagoras was wrong. It is not all numbers. Numbers and mathematics are merely mental models for dealing with some aspects of reality. So it is all fun and joy until someone uses it to build weapons.

@sumitpatel6823 6 ай бұрын

Every kid in India would have solved this question very easily😊

@Kolynogy 6 ай бұрын

When solving for 0 you cant just divide with cos(x) because that Could be 0 .

@farhansadik5423 6 ай бұрын

It could be, but since tanx is already present within the equation, you already know cosx cant be 0 (90 degrees) or else tanx is undefined, which you have to deduce

@mikeschieffer2644 6 ай бұрын

Instead of dividing both sides of the equation sinx = sqrt(3)*cosx by cosx, another way to solve is as follows: sinx - sqrt(3)*cosx = 0. Let A = sqrt(1^2 + (-sqrt(3))^2) = 2 and B = arctan(1/-sqrt(3)) = 5*pi/6. Now instead of sinx - sqrt(3)*cosx = 0 we have Acos(x - B) = 0 or 2cos(x - 5*pi/6) = 0. Divide both sides by 2, take the arccos of both sides, and add 5*pi/6 to both sides. We get 5*pi/6+ pi/2 = 4*pi/3, and 5*pi/6 - pi/2 = pi/3