Question

Find the range of $\theta$ in, $8{\sin }^3\theta -7\sin \theta +\sqrt{3}\cos \theta$

Answer 1

Find the range of $\theta$ in,

$8{\sin }^3\theta -7\sin \theta +\sqrt{3}\cos \theta$

$=8{\sin }^3\theta -6\sin \theta +\sqrt{3}\cos \theta -\sin \theta$

$=-2\sin 3\theta +\sqrt{3}\cos \theta -\sin \theta$

Let, $-1\le \sin 3\theta \le 1$ & $\sqrt{3}\cos \theta -\sin \theta$ range

$\therefore 2\ge -2\sin 3\theta \ge -2$ $=[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}]$

$\therefore -2\ge -2\sin 3\theta \ge 2$ $=[-\sqrt{(\sqrt{3}{)}^2+(-1{)}^2},\sqrt{(\sqrt{3}{)}^2+(-1{)}^2}]$

$=[-2,2]$

Let, $-2\le \sqrt{3}\cos \theta -\sin \theta \le 2$

$-2\le -2\sin 3\theta \le 2$

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$-4\le \sqrt{3}\cos \theta -\sin \theta -2\sin 3\theta \le 4$

($\because$ adding both the range)