cos^2x=sinxcosx

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November 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
November 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 $\cos(x)=0~\&~\cos(x)=\sin(x)~.$
November 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
November 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)
November 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.