Question

${ \sin(x) }^{4} + { \cos(x) }^{4} = \sin(2x) - \frac{1}{2}$
Тригонометрия!​

Answer 1

$\sin^4x+\cos^4x=\sin2x-\dfrac{1}{2} \\ \underbrace{\sin^4x+2\sin^2x\cos^2x+\cos^4x}_{a^2+2ab+b^2=(a+b)^2}=\sin2x-\dfrac{1}{2}+2\sin^22x\cos^22x \\ (\underbrace{\sin^2x+\cos^2x}_{\sin^2x+\cos^2x=1})^2 =\sin2x-\dfrac{1}{2}+\dfrac{\sin^22x}{2} \\ 1=\sin2x-\dfrac{1}{2}+\dfrac{\sin^22x}{2} \\ \sin^22x+2\sin2x=3 \\ -------------------- \\ f(x)=\sin^22x+2\sin2x \\ \sin2x \in [-1; \ 1] \ \Rightarrow \ \sin2x=1, \ \ f(x)_{max}=3 \\ -------------------- \\ \sin2x=1 \\ 2x=\dfrac{\pi}{2}+2\pi k$

$\boxed{x=\dfrac{\pi}{4}+\pi k, \ k \in \mathbb{Z}}$

Ответ: x=π/4+πk, k∈Z