SL 3.8—Solving trig equations

Rearranging: $2\cos x=\sqrt{3}$, so $\cos x=\frac{\sqrt{3}}{2}$.

From the exact values table, $\cos \alpha =\frac{\sqrt{3}}{2}$ gives $\alpha =\frac{\pi }{6}$.

Since $\cos x>0$, solutions lie in Quadrants I and IV. In $[0,2\pi ]$: $x=\frac{\pi }{6}$ and $x=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$.

The domain is $0\le x\le \pi$. Only $x=\frac{\pi }{6}$ lies within this interval; $\frac{11\pi }{6}$ is outside. The answer is $x=\frac{\pi }{6}$.

We need all solutions to $\cos x=\frac{\sqrt{3}}{2}$ in $[0,\pi ]$.

$\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}\ne \frac{\sqrt{3}}{2}$. Quadrant II gives negative cosine, so this is incorrect.

$\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\ne \frac{\sqrt{3}}{2}$. This corresponds to the wrong exact value.

$x=\frac{11\pi }{6}$ lies outside the given domain $[0,\pi ]$, so any option including it is invalid.

$\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$ ✓ and $\frac{\pi }{6}\in [0,\pi ]$ ✓. The answer is $x=\frac{\pi }{6}$.

Let $u=\sin x$. Then $2u^2-u-1=(2u+1)(u-1)=0$, giving $u=-\frac{1}{2}$ or $u=1$.

In $[0°,360°]$, $\sin x=1$ gives $x=90°$. Wait — checking: $\sin 90°=1$ ✓.

Reference angle: $\alpha =30°$. Since $\sin x<0$, solutions are in Quadrants III and IV: $x=180°+30°=210°$ and $x=360°-30°=330°$.

The solutions are $x=90°,210°,330°$. Checking the options, this matches '$x=90°,210°,330°$'.

$(2\sin x+1)(\sin x-1)=0$ gives $\sin x=-\frac{1}{2}$ or $\sin x=1$.

$\sin 270°=-1\ne -\frac{1}{2}$ and $\sin 30°=\frac{1}{2}\ne -\frac{1}{2}$. This option is wrong.

This option is incomplete: it misses $\sin x=1$ (giving $90°$) and the Quadrant IV solution $330°$.

$\sin 150°=\frac{1}{2}$, which does not satisfy either equation, and $\sin 270°=-1\ne -\frac{1}{2}$.

$\sin 90°=1$ ✓, $\sin 210°=-\frac{1}{2}$ ✓, $\sin 330°=-\frac{1}{2}$ ✓. This is the correct answer.