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, $8{t^3} – 8{t^2} + 1 = 0$ Continue reading Solve for $x \in \left[ {0,\frac{\pi }{2}} \right]$

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

Since cos x lies between -1 & 1, the left hand side lies between 4 & 16.

Since sin x lies between -1 & 1 or $sin^8 x $ lies between 0 & 1, the right hand side lies between 2 & 4.

The two sides can become equal only when each one of them is equal to 4.

This happens when cos x of the left hand side is equal to -1 and sin x of the right hand side is equal to 0.

This can only happen when x is an odd multiple of $\pi $.

So, $x=(2n+1)\pi $ where n is any integer.