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Solve for $x \in \left[ {0,\frac{\pi }{2}} \right]$

$8{\sin ^2}x + 2\cos 3x + 6\cos x = 7$

We have, $8(1 - {\cos ^2}x) + 2(4{\cos ^3}x - 3\cos x) + 6\cos x = 7$

$ \Rightarrow 8{\cos ^3}x - 8{\cos ^2}x + 1 = 0$

Let $cos x = t$

So, $8{t^3} - 8{t^2} + 1 = 0$

$t=\frac {1}{2}$ satisfies the equation.

So, $4{t^2}(2t - 1) - 2t(2t - 1) - (2t - 1) = 0$

$ \Rightarrow (2t - 1)(4{t^2} - 2t - 1) = 0$

So, $t = \frac{1}{2},\frac{{1 \pm \sqrt 5 }}{4} = \cos x$

Thus, $x = \frac{\pi }{3},\frac{{\pi }}{5}$ as $x \in \left[ {0,\frac{\pi }{2}} \right]$ in which cos x cannot be negative. $\frac{{1 - \sqrt 5 }}{4}$ solution is rejected.

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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