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
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
so the answer is
\(12-5 \sqrt{5}\)If the big equation were instead
the result would be simply 1