Trigonometric equation

Hello, james_bond!

Quote:

Solve the equation: . $(\sin x+\cos x)\cdot \sin {2x}\:=\:\sqrt{2}$

Square both sides: . $(\sin x + \cos x)^2\sin^2\!2x \:=\:2$

. . . . . . $\left(\sin^2\!x + 2\sin x\cos x + \cos^2\!x\right)\sin^2\!2x \:=\:2$

. . . . . . $(\underbrace{2\sin x\cos x}_{\text{This is }\sin2x} + \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}})\sin^2\!2x \;=\;2$

. . . . . . . . . . . . . . . $(\sin2x + 1)\sin^2\!2x \:=\:2$

. . . . . . . . . . . . . . $\sin^3\!2x + \sin^2\!2x - 2 \:=\:0$

$\text{Factor: }\;\underbrace{(\sin2x-1)}_{\downarrow}\underbrace{\left(\sin^2\!2x + 2\sin2x + 2\right)}_{\text{No real roots}} \:=\:0$

$\text{Then: }\:\sin2x \:=\:1 \quad\Rightarrow\quad2x \:=\:\tfrac{\pi}{2} + 2\pi n \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{4}+ \pi n$

But we find that half of the roots are extraneous.
The equation is satisfied only if $x$ is in Quadrant 1.

Therefore: . $x \;=\;\frac{\pi}{4} + 2\pi n$