Question

решите как можно подробнее 2 задания ​

Answer 1

$sin3x-cos3x=\sqrt{2}sinx\\sin3x-cos3x=\sqrt{2}sin\left ( 3x-\frac{\pi}{4} \right )\\\sqrt{2}sin\left ( 3x-\frac{\pi}{4} \right )=\sqrt{2}sinx\\2sin\left ( x-\frac{\pi}{8} \right )cos\left ( 2x-\frac{\pi}{8} \right )=0\\x=\left \{ \frac{\pi}{8}+\pi k;\frac{5\pi}{16}+\frac{\pi k}{2} \right \},k\in \mathbb{Z}$

$cos\left ( arcsin\frac{1}{5} \right )=\sqrt{1-\frac{1}{25}}=\sqrt{\frac{24}{25}}=\frac{2\sqrt{6}}5{}\\\bigstar \; cos\left ( arcsinx \right )=\sqrt{1-x^2}$