# How to solve the following trig equation

• Dec 15th 2012, 11:26 AM
homeylova223
How to solve the following trig equation
r=a sin(x)

r=a cos(2x)

How would I solve the following equation I know cos(2x)=2cos^2-1,1-sin^2(x)

But I am not sure where to begin.
• Dec 15th 2012, 11:55 AM
jakncoke
Re: How to solve the following trig equation
Set them equal to each other $a sin(x) = a cos(2x)$, the a's cancel, so you get $sin(x) = cos(2x)$.
Now, $cos(2x) = 1 - 2*sin^2(x)$, so $sin(x) = 1-2 sin^2(x)$, so you got $sinx + 2 sin^2(x) = 1$, so $sin(x)[1 + 2sin(x)] = 1$. You can see that at $\frac{3\pi}{2}$ $sin(\frac{3\pi}{2}) = -1$, so $sin(\frac{3\pi}{2})[1 + 2*sin(\frac{3\pi}{2})] = -1*(1 + 2(-1)) = -1*-1 = 1$.
So your solution set would be $\frac{3\pi}{2} + 2\pi * k$ where K is an integer.
• Dec 15th 2012, 11:57 AM
skeeter
Re: How to solve the following trig equation
$a\sin(x) = a\cos(2x)$

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

$2\sin^2{x} + \sin{x} - 1 = 0$

$(2\sin{x} - 1)(\sin{x} + 1) = 0$

can you finish?
• Dec 15th 2012, 03:36 PM
Prove It
Re: How to solve the following trig equation
Let's not forget, you can NOT divide by 0. What should have been done is...

\displaystyle \begin{align*} a\sin{(x)} &= a\cos{(2x)} \\ a\sin{(x)} - a\cos{(2x)} &= 0 \\ a\left[ \sin{(x)} - \cos{(2x)} \right] &= 0 \\ a = 0 \textrm{ or } \sin{(x)} - \cos{(2x)} &= 0 \end{align*}

and then follow the abovementioned methods...
• Dec 15th 2012, 06:40 PM
homeylova223
Re: How to solve the following trig equation