Solving for x, \(\displaystyle{{\sin}^{{{10}}}{x}}+{{\cos}^{{{10}}}{x}}={\frac{{{29}}}{{{16}}}}{{\cos}^{{4}}{2}}{x}\) I had found the

Alexis Alexander

Alexis Alexander

Answered question

2022-04-03

Solving for x, sin10x+cos10x=2916cos42x
I had found the answer as 4x=π2+nπx=π8+nπ4 where nZ. However, I'm not sure whether or not it seems correct.

Answer & Explanation

Luciana Cline

Luciana Cline

Beginner2022-04-04Added 14 answers

We need to sove that
sin8x+cos8xsin6xcos2xsin2xcos4x+sin4xcos4x=2916cos42x
or
(sin4x+cos4x)22sin4xcos4xsin2xcos2x(12sin2xcos2x)+sin4xcos4x=2916cos42x
or
(12sin2xcos2x)22sin4xcos4xsin2xcos2x(12sin2xcos2x)+sin4xcos4x=2916cos42x
=2916cos42x
or
15sin2xcos2x+5sin4xcos4x=2916(1sin22x)2
or
154sin22x+516sin42x=2916(1sin22x)2
Let sin22x=t
Thus
154t+516t2=2916(1t)2
or
24t238t+13=0
which gives t=12 or
sin22x=12
or
cos4x=0
and we got your answer.

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