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I am trying to solve the following equation for x. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm going wrong? Notice that this is "sin squared x" and 3 * "cos squared x"

$\sin^2x = 3\cos^2x$ //Just rewriting the equation again

$1-\cos^2x = 3\cos^2x$ //Using the Pythagorean identities to substitute in for $\sin^2x$

I then add $\cos^2x$ to both sides yielding: $$1 = 4\cos^2x$$

I then divide by $4$ yielding: $$\frac 1 4 = \cos^2x$$

I then take the square root of both sides yielding: $$\frac 1 2 = \cos x$$

Then, I determine that the places where the $\cos x$ is positive $(1/2)$ is $\pi/3$ and $5\pi/3$

The textbook's answer is $\pi/3$ and $2\pi/3$. However the $\cos(2\pi/3)$ is $-(1/2)$ meaning that I must've solved the equation incorrectly. Does anyone see my mistake?

Thanks!

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    $\begingroup$ Divide by $\cos^2 x$, which gives you $\tan^2 x = 3$. $\endgroup$ Commented Nov 8, 2013 at 0:20
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    $\begingroup$ Why do you insist that $\cos x$ must be positive? That doesn't seem to come from anywhere. $\endgroup$ Commented Nov 8, 2013 at 0:21

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You are correct until the square root: This leads to

$$\cos{x} = \pm \frac 1 2$$

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  • $\begingroup$ I see... But Why then is the answer only pi/3 and 2pi/3 ... should the other pi/3's be included as well? $\endgroup$
    – foobar512
    Commented Nov 8, 2013 at 1:01
  • $\begingroup$ @Blakeasd Yes, $5\pi/3$ is also a solution. Is there any restriction on the choice of $x$? $\endgroup$
    – user61527
    Commented Nov 8, 2013 at 1:02
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    $\begingroup$ Unless there is some domain restriction given for the equation, there are four solutions in the principal circle, $ \ 0 \ \le \ x \ < \ 2 \pi \ . $ I suspect that the solver applied the inverse cosine function to $ \ \cos x \ = \ \pm \frac{1}{2} \ $ to obtain only the first and second quadrant solutions. $\endgroup$ Commented Nov 8, 2013 at 1:11
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we know that $\sin^2x=3\cos^2x$ so$$1-\cos^2x=3\cos^2x \Rightarrow 4\cos^2x=1 \Rightarrow \cos^2x=\frac{1}{4} \Rightarrow \cos x=\pm \frac{1}{2}$$ and now we calculate values of $x$ $$\mbox{if} \qquad \cos x=\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(\frac{1}{2}\right)=\frac{\pi}{3},\frac{5\pi}{3}$$ $$\mbox{if} \qquad \cos x=-\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3},\frac{4\pi}{3}$$

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  • $\begingroup$ $\Large\color{red}{ +}$ $\endgroup$
    – M.H
    Commented Dec 8, 2013 at 17:13
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Here is an alternative approach as suggested in the comments by @BennettGardiner. If you simply divide both sides by $\cos^2x$ then the problem will be really easy to solve for you.

\begin{align} &\sin^2(x) = 3\cos^2 (x) \\ &\frac{\sin^2(x)}{\cos^2 (x)} = 3 \\ &\tan^2{x} = 3 \\ &\tan {x} = \pm \sqrt{3} \end{align}

From the last step, you can just $\arctan$ both sides twice, once with $\sqrt{3}$ and another with $-\sqrt{3}$ and you will have both of your answers.

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  • $\begingroup$ You lost a square on the cosine twice. I'm not sure why it is easier with $\tan^2$ than $\cos^2$ $\endgroup$ Commented Nov 12, 2013 at 22:49
  • $\begingroup$ Good catch! Thanks. Personally, I found this route reason because I can just think of it as one term and what follows is just a square and the use of a calculator or a triangle. It doesn't require the use of trig identities or anything. $\endgroup$
    – Jeel Shah
    Commented Nov 12, 2013 at 23:10

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