Solving for $x$, $\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$

Question

$$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$$

• Solve for $x$

I had found the answer as $4x = \dfrac{\pi}{2} + n\pi \implies x = \dfrac{\pi}{8} +\dfrac{n\pi}{4}$ where $n \in \mathbb{Z} $. However, I'm not sure whether or not it seems correct.

Regards

Answer 1

We need to sove that $$\sin^8x+\cos^8x-\sin^6x\cos^2x-\sin^2x\cos^4x+\sin^4x\cos^4x=\frac{29}{16}\cos^42x$$ or $$(\sin^4x+\cos^4x)^2-2\sin^4x\cos^4x-\sin^2x\cos^2x(1-2\sin^2x\cos^2x)+\sin^4x\cos^4x=\frac{29}{16}\cos^42x$$ or $$(1-2\sin^2x\cos^2x)^2-2\sin^4x\cos^4x-\sin^2x\cos^2x(1-2\sin^2x\cos^2x)+\sin^4x\cos^4x=\frac{29}{16}\cos^42x$$ or $$1-5\sin^2x\cos^2x+5\sin^4x\cos^4x=\frac{29}{16}(1-\sin^22x)^2$$ or $$1-\frac{5}{4}\sin^22x+\frac{5}{16}\sin^42x=\frac{29}{16}(1-\sin^22x)^2$$ Let $\sin^22x=t$.

Thus, $$1-\frac{5}{4}t+\frac{5}{16}t^2=\frac{29}{16}(1-t)^2$$ or $$24t^2-38t+13=0,$$ which gives $t=\frac{1}{2}$ or $$\sin^22x=\frac{1}{2}$$ or $$\cos4x=0$$ and we got your answer.

Answer 2

The LHS can be rewritten

$$\left(\frac{1-\cos 2x}2\right)^5+\left(\frac{1+\cos 2x}2\right)^5=\frac{5\cos^42x+10\cos^22x+1}{16},$$ leading to the biquadratic equation

$$24\cos^42x-10\cos^22x-1=0.$$

There are only two real roots, $\cos 2x=\pm\dfrac1{\sqrt 2}$. The rest is easy.

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The resolution strategy was to express both members in terms of $\cos2x$ (using the double angle formula), and luckily all the odd degree terms cancel out, leaving a biquadratic equation.

Answer 3

Using $$ (a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5),(a-b)^5=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5)$$ and $$ \cos^2x=\frac{1\cos(2x)}{2},\sin^2x=\frac{1-\cos(2x)}{2}$$ one has \begin{eqnarray} \cos^{10}x+\sin^{10}x&=&\frac1{2^5}(1+\cos(2x))^5+\frac1{2^5}(1-\cos(2x))^5\\ &=&\frac{1}{16}(1+10\cos^2(2x)+5\cos^4(2x)) \end{eqnarray} and hence the equation becomes $$ 1+10\cos^2(2x)+5\cos^4(2x) = 29\cos^4(2x)$$ or $$ 24\cos^4(2x)-10\cos^2(2x)-1=0. $$ or $$ (12\cos^2(2x)+1)(2\cos^2(2x)-1)=0.$$ So $$2\cos^2(2x)-1=0$$ or $$ \cos(4x)=0$$ which gives $$4x=n\pi+\frac{\pi}{2}$$ or $$ x=\frac{n\pi}{4}+\frac{\pi}{8}, n\in\mathbb{Z}. $$

Answer 4

Using trigonometric identities: $$\begin{align}\sin^{10}{x}+\cos^{10}{x}&=\frac{29}{16}\cos^4{2x} \Rightarrow \\ (\sin^2x+\cos^2x)^5-5\sin^2x\cos ^2x(\sin^6x+\cos^6x)-10\sin^4x\cos^4x&=\frac{29}{16}\cos^4{2x}\Rightarrow\\ 1-\frac{5}{4}\sin^22x((\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x)-\frac{10}{16}\sin^42x&=\frac{29}{16}\cos^4{2x}\Rightarrow\\ 1-\frac{5}{4}\sin^22x(1-\frac{3}{4}\sin^22x)-\frac{10}{16}\sin^42x&=\frac{29}{16}(1-\sin^2{2x})^2\Rightarrow\\ 16-20\sin^22x+15\sin^42x-10\sin^42x&=29-58\sin^22x+29\sin^42x\Rightarrow \\ 24\sin^42x-38\sin^22x+13&=0 \Rightarrow \\ \sin^22x\ne \frac{13}{12} \ \text{or} \ \sin^22x&=\frac 12 \Rightarrow \\ \sin 2x&=\pm \frac{1}{\sqrt{2}} \Rightarrow \\ 2x&=\frac{\pi}{4}+\frac{\pi n}{2}, n \in \mathbb{Z} \Rightarrow \\ x&=\frac{\pi}{8}+\frac{\pi n}{4}, n \in \mathbb{Z}.\end{align}$$