0
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.