1. ## Trig equation

Hi there,

I'm having a little trouble solving the following equation for $\displaystyle -\pi <x\leq 2 \pi$:

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

I've rearranged it into the form:

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

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

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

2. Originally Posted by Stroodle
Hi there,

I'm having a little trouble solving the following equation for $\displaystyle -\pi <x\leq 2 \pi$:

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

I've rearranged it into the form:

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

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

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

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

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

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

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

$\displaystyle 0 = u^2 - u + 1$

note that $\displaystyle b^2-4ac < 0$ ... therefore, this factor of your original equation has no solution.

3. ahh, of course

Thanks!

4. Hello, Stroodle!

Another approach . . .

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

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

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

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

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

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

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

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

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