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$3\sin^2x=\cos^2x;$ $0\leq x\leq 2\pi$ Solve for $x$:

I honestly have no idea how to start this. Considering I'm going to get a number, I am clueless. I have learned about $\sin$ and $\cos$ but I do not know how to approach this problem. If anyone can go step-by-step with hints. That would be greatly appreciated.

EDIT:
$$3\sin^2x=1-\sin^2x$$ $$4\sin^2x=1$$ $$\sin^2x=\frac{1}{4}$$ $$\sqrt{\sin^2x}=\sqrt{\frac{1}{4}}$$ $$\sin x=\pm \left(\frac{1}{2}\right)$$

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3 Answers 3

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Hint: $\cos^2 x=1-\sin^2 x$.${}{}{}{}{}{}{}$

Substitute. We get after some simplification $4\sin^2 x=1$. Can you finish from here?

Added: You just need to find $\sin^2 x$, then $\sin x$. You should get $\sin x=\pm\frac{1}{2}$. Then identify the angles from your knowledge about "special angles." One of the angles will turn out to be $\frac{\pi}{6}$, the good old $30^\circ$ angle. There are $3$ others.

Remark: The way you started is fine too, you got $3(1-\cos^2 x)=\cos^2 x$. Now we need to "isolate" $\cos^2 x$. It is easiest to multiply through by $3$ on the left, getting $3-3\cos^2 x=\cos^2 x$. Bring all the $\cos^2 x$ terms to one side. We get $3=4\cos^2 x$, which I prefer to write as $4\cos^2 x=3$. So we get $\cos^2 x=\frac{3}{4}$.

Take the square roots. We get $$\cos x=\pm \frac{\sqrt{3}}{2}.$$

Following the hint given at the start happens to be a little easier, same principles, nicer numbers.

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  • $\begingroup$ If we measure angles in degrees, the sine is $-1/2$ at $210$ degrees and $330$ degrees. this is because the sine curve between $180$ and $360$ ($\pi$ and $2\pi$) is just the same shape as between $0$ and $180$, but flipped over (negative). I don't mind answering questions, but we should delete most of the comments, the system gets upset. $\endgroup$
    – André Nicolas
    Jul 16, 2012 at 22:47
  • $\begingroup$ The sine at $1/2$ is not $210$ or $330$. The result is that $\sin(210^\circ)=-1/2$, and $\sin(330^\circ)=-1/2$. $\endgroup$
    – André Nicolas
    Jul 16, 2012 at 23:00
  • $\begingroup$ @AustinBroussard: I am sure there are lots of better one. But you will find some pictures here and lots more. Your book will also have pictures. $\endgroup$
    – André Nicolas
    Jul 16, 2012 at 23:04
  • $\begingroup$ Ah, speaking of $\sin (210^\circ )=-\dfrac{1}{2}$ how would you solve that? As in, $\sin x=\dfrac{1}{2}$? I don't know if you understand what I'm trying to ask or not. $\endgroup$
    – Austin Broussard
    Jul 16, 2012 at 23:12
  • $\begingroup$ The shape of the sine curve tells you that. Since $\sin (30^\circ)=\frac{1}{2}$, we must have $\sin(30^\circ+180\circ)=-\frac{1}{2}$. Knowing the trig functions between $0$ and $90$ lets you figure them out everywhere else. Look at the picture of sine, look at what has been written, look at the picture again, practice a bit. It will click. $\endgroup$
    – André Nicolas
    Jul 16, 2012 at 23:21
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$$ 3\sin^2 x = \cos^2 x $$ $$ 3\sin^2 x = 1 - \sin^2 x $$ $$ 3u^2 = 1 - u $$ That's a quadratic equation. After you solve it for $u$, write $\sin x = \text{whatever you got for }u$ and then find $x$.

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You can also do this. Divide out to get $$\tan^2(x) = \left({\sin(x)\over \cos(x)}\right)^2 = {1\over 3}.$$ This gives $\tan(x) = \pm 1/\sqrt{3}$.

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