Question

2sin(0,5п+2x)
———————
ctgx+tg(п-x)

П
Х= —
3

Answer 1

$\frac{2\sin\left(\frac\pi2+2x\right)}{ctgx+tg\left(\pi-x\right)}=\frac{2\cos2x}{ctgx-tgx}=2\cos2x:\left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)=2\cos2x:\left(\frac{\cos^2x-\sin^2x}{\sin x\cos x}\right)=\\\\=2\cos2x:\frac{\cos2x}{\sin x\cos x}=2\cos2x\cdot\frac{\sin x\cos x}{\cos2x}=2\sin x\cos x=\sin2x\\\\x=\frac\pi3:\\\sin2x=\sin\frac{2\pi}3=\frac{\sqrt3}2$