# Solving $\sin^{2015}x+\cos^{2015}x=\frac12$

Find all the roots of $$\sin^{2015}x+\cos^{2015}x=\frac12\tag{1}$$

I'm a high school student, and this is my homework. This's my try:

Let $\displaystyle t=\tan \frac x2\Rightarrow \sin x=\frac{2t}{1+t^2}, \ \ \cos x=\frac{1-t^2}{1+t^2}$

We'll have $\displaystyle\color{Red}{(1)} \Leftrightarrow \frac{(2t)^{2015}+\sum\limits_{k=0}^{2015}\binom{2015}k\cdot \left ( -1 \right )^{2015-k}\cdot t^{2k }}{\left (1+t^2 \right )^{2015}}=\frac12$

• Hint: you need either $\sin x$ or $\cos x$ very close to $1$, else the left hand side is very close to $0$ (and can't equal $\frac 12$). – vadim123 Aug 10 '15 at 2:02
• You can just neglect one of the terms and work only with the other one like solve $cos(x)^{2015}=\frac{1}{2}$. Do the same for sin(x) and you get all the answers. It works because the other number will be a number close to 0 which will almost banish when taking the 2015th power. – Martín Forsberg Conde Aug 10 '15 at 2:38
• $x\simeq0.02622800777527157466780627531660\ldots$ – Lucian Aug 10 '15 at 2:57
• $x\simeq\arccos\dfrac1{\sqrt[2015]2}$ or $x\simeq\arcsin\dfrac1{\sqrt[2015]2}$ – Lucian Aug 10 '15 at 3:02
To get an approximation, note that either $\cos x$ or $\sin x$ must be very close to $1$. The other will be tiny, so we ignore it. Looking for the root just above $0$, you can use the Taylor series for the cosine. $(1-\frac {x^2}2)^{2015}=\frac 12$ This is a quadratic in $x$