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Given a trigonometric equation : $$\sin(2x)^{2016} + \cos(2x)^{2016} = 1$$

How many solutions does this equation has given x belongs in the interval [0,2$\pi$)?

I thought the obvious one would be when $\sin(2x)$ is zero and $\cos(2x)$ is 1 as 0 and 1 are not affected by the exponent 2016. Also I can take -1 as value also as exponent is even. So I took all those values and got values like 2x = {0, $\pi$/2, $\pi$, ...etc}.

Are there any solutions apart from these? How many total solutions there ?

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2 Answers 2

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No there are no solutions apart from when one of them is 0. It can be seen as follows, $$ \sin(2x)^{2016}+\cos(2x)^{2016}=(\sin(2x)^2+\cos(2x)^2)^{1008}-S $$ As S consists of terms of the form ${1008 \choose n }\sin(2x)^{2n}\cos(2x)^{2(1008-n)}$, each term is positive and hence $S\geq0$. Therefore, $$ \sin(2x)^{2016}+\cos(2x)^{2016}\leq1$$ Equality occurs only when one of them is 0.

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If x isn't one of the points you mentioned then note this: for $n >1$

$|\sin(x)| < 1 \rightarrow |\sin(x)|^n < |\sin(x)|$. The same applies to cosinus. We can conclude:

$\sin(x)^{2016}+\cos(x)^{2016} < \sin(x)^2 + \cos(x)^2 = 1$

So you have all the solutions in R. (this works fine since 2016 is even, for oneven power we would need few more steps in the proof).

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