Q.Solve the equation $\cos x = \dfrac{1}{2}$.

The quartic in $\sin x$ is a perfect square that forces $\sin^2x=1$, which has exactly two solutions in $[0,2\pi]$.

1. The equation is $\sin^4x-2\sin^2x+1=0$.
2. Let $t=\sin^2x$. Then $t^2-2t+1=0\Rightarrow(t-1)^2=0\Rightarrow t=1$.
3. So $\sin^2x=1\Rightarrow\sin x=\pm1$. …

$\cos^2\theta=\cos^2\alpha$ is equivalent to $\cos\theta = \pm\cos\alpha$, whose combined general solution is $\theta = n\pi\pm\alpha$.

$\cos^2\theta = \cos^2\alpha \implies \cos^2\theta - \cos^2\alpha = 0$

$(\cos\theta - \cos\alpha)(\cos\theta+\cos\alpha) = 0$

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Recognising $\sin^4x-2\sin^2x+1$ as a perfect square in $\sin^2x$ reduces the equation to $\cos x=0$, which has 2 solutions in $[0,2\pi]$.

1. The equation is $\sin^4x-2\sin^2x+1=0$.
2. This factors as a perfect square: $\sin^4x-2\sin^2x+1=(\sin^2x-1)^2$.
3. So $(\sin^2x-1)^2=0\Rightarrow\sin^2x=1\Rightarrow\cos^2x=1-\sin^2x=0\Rightarrow\cos x=0$. …

The general solution of $\tan\theta = 1$ is $\theta = n\pi + \pi/4$; substituting $\theta = 3x$ and solving for $x$ gives $\dfrac{n\pi}{3}+\dfrac{\pi}{12}$.

The general solution of $\tan\theta = \tan\alpha$ is:

$\theta = n\pi + \alpha, \quad n \in \mathbb{Z}$

Here $\tan3x = 1 = \tan\dfrac{\pi}{4}$, so with $\theta = 3x$ and $\alpha = \pi/4$: …

Use the general solution of $\tan\theta = 0$.

$\tan 2x = 0$

The general solution of $\tan\theta = 0$ is $\theta = n\pi,\ n\in\mathbb{Z}$.

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$\sin x = \dfrac{\sqrt3}{2}$ has general solution $x=n\pi+(-1)^n\dfrac{\pi}{3}$.

We know $\sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2}$, so the equation $\sin x = \sin\dfrac{\pi}{3}$ has the standard general-solution form …