# Trig equation

• Mar 2nd 2010, 02:31 PM
Stroodle
Trig equation
Hi there,

I'm having a little trouble solving the following equation for $-\pi :

$sin^2(x)cos^3(x)=cos(x)$

I've rearranged it into the form:

$cos(x)\left [sin^2(x)cos^2(x)-1\right ]=0$

So where $cos(x)=0,\ \ \ x= \pm\frac{\pi}{2}$ and $\frac{3\pi}{2}$

But I'm just not sure how to find $x$ where $sin^2(x)cos^2(x)=1$

• Mar 2nd 2010, 03:51 PM
skeeter
Quote:

Originally Posted by Stroodle
Hi there,

I'm having a little trouble solving the following equation for $-\pi :

$sin^2(x)cos^3(x)=cos(x)$

I've rearranged it into the form:

$cos(x)\left [sin^2(x)cos^2(x)-1\right ]=0$

So where $cos(x)=0,\ \ \ x= \pm\frac{\pi}{2}$ and $\frac{3\pi}{2}$

But I'm just not sure how to find $x$ where $sin^2(x)cos^2(x)=1$

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

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

$0 = \sin^4{x} - \sin^2{x} + 1$

let $u = \sin^2{x}$ ...

$0 = u^2 - u + 1$

note that $b^2-4ac < 0$ ... therefore, this factor of your original equation has no solution.
• Mar 2nd 2010, 04:17 PM
Stroodle
ahh, of course :)

Thanks!
• Mar 2nd 2010, 04:23 PM
Soroban
Hello, Stroodle!

Another approach . . .

Quote:

Solvie the following equation for $-\pi < x\leq 2 \pi\!:\;\;\sin^2\!x\cos^3\!x\:=\L\cos x$

I've rearranged it into the form: $\cos(x)\left[\sin^2\!x\cos^2\!x-1\right ]\:=\:0$

So: $\cos x\,=\,0,\;x\:=\: \pm\frac{\pi}{2}\,\text{ and }\,\frac{3\pi}{2}$

But I'm not sure how to solve: $\sin^2x\cos^2x \:=\:1$

Take square roots: . $\sin x\cos x \:=\:\pm1$

$\text{Multiply by 2: }\;\underbrace{2\sin x\cos x} \:=\:2$

$\text{Then we have: }\quad\;\;\overbrace{\sin2x} \:=\:2$ .[1]

But $|\sin2x| \,\leq\,1$

. . Therefore, [1] has no real solutions.