Solving trigonometric equations to the fourth power.

Question

$$\sin^4(x)-\sin^2(x)=0$$

My work:

Let $t=\sin^2(x)$

Rewrite the original equation as: $t^2-t$

Factor: $t(t-1)$

$t=1$, $t=0$

What do I do from here?

Answer 1

Change variables back from $t$ to $x$, then solve each equation separately:

\begin{align*} t = 1 &\iff \sin^2 x = 1 \\ &\iff \sin x = \pm 1 \\ &\iff x = \frac{\pi}{2} + n\pi ~~~~~\text{where } n \in \mathbb Z \end{align*}

Likewise:

\begin{align*} t = 0 &\iff \sin^2 x = 0 \\ &\iff \sin x = 0 \\ &\iff x = n\pi ~~~~~\text{where } n \in \mathbb Z \end{align*}

Combining the two above cases, we obtain: $$ x = \frac{n\pi}{2} ~~~~~\text{where } n \in \mathbb Z $$

Answer 2

You have almost done it. From there you get $$\sin^2x=0$$ or $$\sin^2 x=1$$. Hence Either $\sin x=1$, $\sin x=-1$ or $\sin x=0$. So you get $x = k\pi$, $ k \in Z$ or $ x=k \pi \pm \dfrac{\pi}{2}$ where $ k \in Z$

Answer 3

I prefer this approach:

$$\sin^4(x) - \sin^2(x) = \sin^2(x)(\sin^2(x)-1) = \sin^2(x)(-\cos^2(x)).$$

The last equality follows from Pythagorean identity. So then all you have to solve is

$$ 0 = \sin^2(x)\cos^2(x).$$

Or if you want to use double angle formula,

$$ 0 = \frac{1}{4}\sin^2(2x).$$

Answer 4

Why bother with substitution? Just factor it directly

$$ 0 = (\sin x +1)(\sin x - 1)\sin^2 x $$ and solve the equations resulting from setting each distinct factor equal to zero:

$$ \sin x +1 =0 \iff x=-\frac{\pi}{2}+2k\pi $$

$$ \sin x -1 =0 \iff x=\frac{\pi}{2}+2k\pi $$

$$ \sin x =0 \iff x=k\pi $$

Putting these together, the product vanishes precisely when

$$x=\boxed{\dfrac{k\pi}{2}}$$ for integral $k$.