Question

If $2\sin x=\sqrt{3}$, then evaluate $3\cos x-4{\cos }^{3}x$.

Answer 1

$2\sin x=\sqrt{3}$

$\sin x=\frac{\sqrt{3}}{2}$

$\sin x=\sin 60$

$x={60}^{\circ }$
Now, $3\cos x-4{\cos }^{3}x$.

= $3\cos 60-4{\cos }^{3}60$

= $3\left(\frac{1}{2}\right)-4(\frac{1}{2}{)}^3$

= $\frac{3}{2}-\frac{1}{2}$

= $1$