Studying for a trig exam. What tips and key identities should I remember?

Question

I have the following equation:

$$\cos(2x)-\sin(x)=0$$

I'm not sure how to tackle it, I'm very inexperienced with trig identities.

I'm having an exam coming up with equations like these and I want to prepare. I would love some tips and key identities to remember.

Thanks!

Answer 1

The general strategy consists in arriving at one of the standard trigonometric equations: $$\sin x=\sin\alpha,\qquad \cos x=\cos\alpha,\qquad\tan x=\tan\alpha.$$ The solutions are well-known. They are, respectively: $$ \begin{cases}x\equiv \alpha&\bmod 2\pi\\ x\equiv\pi-\alpha&\bmod2\pi\end{cases} \qquad \begin{cases}x\equiv \alpha&\bmod 2\pi\\ x\equiv-\alpha&\bmod2\pi\end{cases} \qquad x\equiv \alpha\mod \pi. $$

Here, you can have two ways:

• either you write $\;\sin x=\cos(\frac\pi 2-x)$ and you get an equation of the second type: $$\cos 2x=\cos(\tfrac\pi 2-x)\iff 2x\equiv \pm\bigl(\tfrac\pi2-x\bigr)\mod2\pi\iff\dotsm.$$
• or you use the duplication formula: $\;\cos 2x=1-2\sin^2x$, so that you obtain a quadratic equation in $\sin x$: $$2\sin^2x+\sin x-1=0,$$ and you solve this quadratic equation (roots $-1$ and $\frac12$), then solve for $x$.

Answer 2

One tip which helps with some, though by no means all, such equations is to think about graphing things. In this case, the graphs $y=\cos2x$ and $y=\sin x$ are fairly simple to sketch. The sine curve starts at $0$, goes up to $1$ at $x=\pi/2$, back to $0$ at $\pi$, down to $-1$ at $3\pi/2$, and back to $0$ at $2\pi$, after which it repeats. The cosine curve, on the other hand, starts at $1$, goes down to $-1$ at $\pi/2$ (because $\cos(2\pi/2)=\cos\pi=-1$), back to $1$ at $\pi$, down again to $-1$ at $3\pi/2$, and back to $1$ at $2\pi$, after which it also repeats. If you draw this, it should be obvious that $x=3\pi/2$ is one solution to the equation $\cos2x=\sin x$, and that there are two other solutions, one that's a bit to the right of $x=0$ and one that's a similar bit to the left of $x=\pi$. You might now guess (and then verify) that those values are $x=\pi/6$ and $x=\pi-\pi/6=5\pi/6$. So the complete solution set consists of the translates, by integer multiples of $2\pi$, of $3\pi/2$, $\pi/6$, and $5\pi/6$.

Note, this may not be the best approach to take on the test itself, where answering questions quickly is a concern. (Sketches can sometimes also be misleading. In particular, it's possible to misdraw things to make it look like there are three solutions near $x=3\pi/2$.) But if your goal is to become facile at working with trigonometric functions, learning to think in terms of how things look is a useful skill.

Answer 3

There are lots of ways to solve trig problems, and unfortunately there isn't a single sure fire way to solve all of them. For your particular problem, there is a nice solution using the identity $\cos{\theta}=\sin{(\frac{\pi}{2}-\theta)}$.

Using this, your equation can be rewritten as: $\sin{(\frac{\pi}{2}-2x)}=\sin{x}$.

Setting the angles equal to each other gives $\frac{\pi}{2}-2x=x$ which can be solved to get $x=\frac{\pi}{6}$

This, of course, is only one of the solutions to this equation. I'll leave it as a fun brainteaser for you to figure out all the other ones.