# Trigonometric equation

• Dec 30th 2008, 02:23 AM
james_bond
Trigonometric equation
Solve the equation:
$\displaystyle (\sin x+\cos x)\cdot \sin {2x}=\sqrt{2}$.
Or if you like the equivavalent:
$\displaystyle \sin x\cos x\cdot(\sin x+\cos x)=\frac{\sqrt{2}}2$.
• Dec 30th 2008, 03:26 AM
Skalkaz
Dear james_bond,

or if you like the equivavalent:
$\displaystyle [(1/\sqrt{2})sin(x)+(1/\sqrt{2})cos(x)]sin(2x)=1$

hint: in [] brackets there is sin(a+b) with suitable a and b.
• Dec 30th 2008, 06:13 AM
Soroban
Hello, james_bond!

Quote:

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

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

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

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

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

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

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

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

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

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