## Solve \(\cos ^{2} x+\cos x-1=0 ?\)

Answer 1:

Given \(\cos x+\cos ^{2} x=1 \ldots \ldots . .(1)\)

\(\Rightarrow \cos x=1-\cos ^{2} x\)

\(\Rightarrow \cos x=\sin ^{2} x\)

Again from (1)

\(\cos ^{2} x+\cos x-1=0\)

\(\begin{aligned}

&\text { And } \cos x=\frac{1}{2}\left(-1+\sqrt{1^{2}-4 \cdot 1 \cdot(-1)}\right) \\

&\Rightarrow \cos x=\frac{1}{2}(-1+\sqrt{5})

\end{aligned}\)

other root

\(\cos x=\frac{1}{2}(-1-\sqrt{5})<-1 \rightarrow \text { not possible }\)So

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When the given expression

Answer 2:

\(12-5 \sqrt{5}\)

Explanation:

Solving

and

\(\sin x=\sqrt{\frac{1}{2}(-1+\sqrt{5})}\)Putting this results into the big equation

\(\sin ^{12} x+\cdots+\sin ^{6} x\) we obtain the answer.

Example

\(\left(\sqrt{\frac{1}{2}(-1+\sqrt{5})}\right)^{16}=\frac{1}{2}(47-21 \sqrt{5})\)so the answer is

\(12-5 \sqrt{5}\)If the big equation were instead

\(\sin ^{12} x+3 \sin ^{10} x+3 \sin ^{8} x+\sin ^{6} x\)the result would be simply 1