Giải bài tập 1.63 trang 30 STB toán 11 tập 1 kết nối

a) Ta có sin 5x + cos 5x = – 1

$\Leftrightarrow \sqrt{2}sin(5x+\frac{\pi}{4})=-1$

$\Leftrightarrow sin(5x+\frac{\pi}{4})=-\frac{1}{\sqrt{2}}$

$\Leftrightarrow sin(5x+\frac{\pi}{4})=sin(-\frac{\pi}{4})$

$\Leftrightarrow 5x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi$ hoặc $5x+\frac{\pi}{4}=\pi-(-\frac{\pi}{4})+k2\pi$

$ \Leftrightarrow x=-\frac{\pi}{10}+k\frac{2\pi}{5}$ hoặc $x=\frac{\pi}{5}+k\frac{2\pi}{5} (k \in \mathbb{Z})$

b) Ta có cos 3x – cos 5x = sin x

$\Leftrightarrow -2sin\frac{3x+5x}{2}sin\frac{3x-5x}{2}=sinx$

$\Leftrightarrow -2sin4xsin(-x)=sinx$

$\Leftrightarrow -2sin4xsinx-sinx=0$

$\Leftrightarrow sin(2sin4x-1)=0$

$\Leftrightarrow sinx=0$ hoặc $sin4x=\frac{1}{2}$

+ Với sin x = 0 ta được $x=k\pi (k \in \mathbb{Z}$

+ Với $sin4x=\frac{1}{2} \Leftrightarrow sin4x=sin(\frac{\pi}{6})$

$\Leftrightarrow 4x=\frac{\pi}{6}+k2\pi$ hoặc $4x=\pi-\frac{\pi}{6}+k2\pi (k\in \mathbb{Z})$

$\Leftrightarrow x=\frac{\pi}{24}+k\frac{\pi}{2}$ hoặc $x=\frac{5\pi}{24}+k\frac{\pi}{2}(k\in \mathbb{Z})$

c) Ta có: $2cos^{2}x+cos2x=2$

$\Leftrightarrow (2cos^{2}x-1)+cos2x=1$

$\Leftrightarrow cos2x +cos2x=1$

$\Leftrightarrow 2cos2x=1$

$\Leftrightarrow cos2x=\frac{1}{2}$

$\Leftrightarrow cos2x=cos\frac{\pi}{3}$

$\Leftrightarrow 2x =\pm \frac{\pi}{3}+k2\pi (k \in \mathbb{Z})$

$\Leftrightarrow x = \pm \frac{\pi}{6} + k\pi (k \in \mathbb{Z})$

d) Ta có $sin^{4}x+cos^{4}x=\frac{1}{2}sin^{2}2x$

$\Leftrightarrow (sin^{2}x+cos^{2}x)^{2}-2sin^{2}xcos^{2}x=\frac{1}{2}sin^{2}2x$

$\Leftrightarrow 1-\frac{1}{2}(2sinx cosx)^{2}=\frac{1}{2}sin^{2}2x$

$ \Leftrightarrow 1-\frac{1}{2}sin^{2}2x=\frac{1}{2}sin^{2}2x $

$\Leftrightarrow sin^{2}2x=1 $

$\Leftrightarrow cos2x = 0 $ (do $sin^{2}2x + cos^{2}2x=1$)

$\Leftrightarrow 2x =\frac{\pi}{2}+k\pi (k \in \mathbb{Z}) $

$\Leftrightarrow x=\frac{\pi}{4}+k\frac{\pi}{2} (k \in \mathbb{Z})$