Question

Решить уравнение 1)√2cos(pi/4 + x) - cosx=1 2)√2sin(pi/4 - x/2) + sin(x/2)=1

Answer 1

$\sqrt{2} cos( \frac{ \pi }{4} +x)-cosx=1 \sqrt{2} (cos \frac{ \pi }{4}cosx-sin \frac{ \pi }{4}sinx )-cosx=1$
$\sqrt{2} ( \frac{ \sqrt{2} }{2}cosx- \frac{ \sqrt{2} }{2} sinx )-cosx=1$$cosx-sinx-cosx=1$
-sinx=1
$x=- \frac{ \pi }{2} =2 \pi n$

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