# 3 Trigonometric Equations

• Nov 26th 2009, 09:30 AM
faris92
2 Trigonometric Equations
I have next week an exam and I don't know how to solve the following 2 equations. If anyone knows how to solve them, please, help me.

sin2(4x) – sin2(2x) = 0
cos(2x) = cos(x) – sin(x)

The teacher gave me the solutions but I don't know how to get them. Here are the solutions:
http://img442.imageshack.us/img442/6461/67701671.png
• Nov 26th 2009, 11:09 AM
I-Think
I'll handle the 1st question,
$sin^2(4x)-sin^2(2x)=0$

$(2sin2xcos2x)^2=(2sinxcosx)^2$

$sin2xcos2x=sinxcosx$

$2sinxcosx(cos^2x-sin^2x)=sinxcosx$

$cos^2x-sin^2x=\frac{1}{2}$

$2cos^2x-1=\frac{1}{2}$

$cos^2x=\frac{3}{4}$

$cosx=\pm{\frac{\sqrt{3}}{2}}$

Values of x that satisfy this equation are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$. And as the $cos(x)$ function is periodic, as a consequence all integer multiples of those functions are your solution.
• Nov 28th 2009, 11:14 AM
rainer
Here's a solution to the second one.

I've doublechecked this and am pretty sure I haven't made any errors.

However, the answers your teacher gave you don't work when I plug them in.

$\cos{2x}=\cos{x}-\sin{x}$

$1-2\sin^2{x}=\cos{x}-\sin{x}$

$(1-2\sin^2{x})^2=(\cos{x}-\sin{x})^2$

$1-4\sin^2{x}+4\sin^4{x}=1-2\cos{x}\sin{x}$

$4\sin^4{x}-4\sin^2{x}=-2\cos{x}\sin{x}$

$4\sin^2{x}(\sin^2{x}-1)=-2\cos{x}\sin{x}$

$4\sin^2{x}(-\cos^2{x})=-2\cos{x}\sin{x}$

$-4\sin^2{x}\cos^2{x}=-2\cos{x}\sin{x}$

$2\sin{x}\cos{x}=1$

$\sin{2x}=1$

$x=\frac{\pi}{4}$ and $x=\frac{\pi}{4}+\pi$ satisfy this equation.