# cos^2x=sinxcosx

• Nov 18th 2012, 08:04 AM
Furyan
cos^2x=sinxcosx
Hello

I am struggling with a simple trig equation, again.

The question is to solve, on the interval $-\pi \leq x \leq \pi$,

$\cos^2x = sinxcosx$

Using my graphing calculator I get four solutions.

Dividing through by $cos^2x$ I get:

$tanx = 1$, for which there are only two solutions. Two of the four I should have.

I have come across this before, where decreasing the power reduces the number of solutions, which makes sense, although I don't understand it. I'd by grateful if someone would explain where I'm going wrong.

Thank you
• Nov 18th 2012, 08:26 AM
Plato
Re: cos^2x=sinxcosx
Quote:

Originally Posted by Furyan
The question is to solve, on the interval $-\pi \leq x \leq \pi$,
$\cos^2(x) = \sin(x)\cos(x)$

That can be written as $\cos(x)(\cos(x)-\sin(x))=0$.

Now solve these two $\cos(x)=0~\&~\cos(x)=\sin(x)~.$
• Nov 18th 2012, 08:28 AM
topsquark
Re: cos^2x=sinxcosx
Quote:

Originally Posted by Furyan
Hello

I am struggling with a simple trig equation, again.

The question is to solve, on the interval $-\pi \leq x \leq \pi$,

$\cos^2x = sinxcosx$

Using my graphing calculator I get four solutions.

Dividing through by $cos^2x$ I get:

$tanx = 1$, for which there are only two solutions. Two of the four I should have.

I have come across this before, where decreasing the power reduces the number of solutions, which makes sense, although I don't understand it. I'd by grateful if someone would explain where I'm going wrong.

Thank you

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

Factoring
$[cos(x)] \cdot cos(x) = [cos(x)] \cdot sin(x)$

Notice that there will be solutions when
cos(x) = 0
and when
cos(x) = sin(x)

-Dan
• Nov 18th 2012, 08:46 AM
Furyan
Re: cos^2x=sinxcosx
Hello Plato and topsquark,

Thank you both very much. I actually understand that now and will look out for it in the future.(Bow)
• Nov 18th 2012, 05:18 PM
Prove It
Re: cos^2x=sinxcosx
Quote:

Originally Posted by Furyan
Hello

I am struggling with a simple trig equation, again.

The question is to solve, on the interval $-\pi \leq x \leq \pi$,

$\cos^2x = sinxcosx$

Using my graphing calculator I get four solutions.

Dividing through by $cos^2x$ I get:

$tanx = 1$, for which there are only two solutions. Two of the four I should have.

I have come across this before, where decreasing the power reduces the number of solutions, which makes sense, although I don't understand it. I'd by grateful if someone would explain where I'm going wrong.

Thank you

The reason you ended up with less solutions than you should have is because \displaystyle \begin{align*} \cos^2{x} \end{align*} CAN equal 0. You can not divide by 0.