Need help with a Trig equation

**daunder** · April 26th 2009, 05:48 AM

$2sin\theta cos\theta-cos\theta=0$

This is what I did

First I took $cos\theta$ to other side resulting in $2sin\theta cos\theta=cos\theta$
Then I divided by $cos\theta$ resulting in $2sin\theta =1$ then I divided by 2 $sin\theta = \frac{1}{2}$
I am sure it's incorrect.. Can someone please help me

**running-gag** · April 26th 2009, 05:53 AM

Hi

Dividing by $\cos\theta$ makes you lose solutions.

You must factor.

$2\sin\theta \cos\theta-\cos\theta=0$

$\cos\theta (2\sin\theta-1)=0$

leading to $\cos\theta = 0$ or $\sin\theta = \frac{1}{2}$

**e^(i*pi)** · April 26th 2009, 05:58 AM

Originally Posted by daunder

$2sin\theta cos\theta-cos\theta=0$

This is what I did

First I took $cos\theta$ to other side resulting in $2sin\theta cos\theta=cos\theta$
Then I divided by $cos\theta$ resulting in $2sin\theta =1$ then I divided by 2 $sin\theta = \frac{1}{2}$
I am sure it's incorrect.. Can someone please help me

That would give you some solutions but not all of them. You can get them all by factorising:

$cos(\theta)(2sin(\theta)-1) = 0$

Then you can solve $cos(\theta) = 0$ to give $\theta = \frac{\pi}{2} \text { , } \frac{3\pi}{2}$

Then solve $2sin(\theta)-1 = 0$

$\theta = \frac{\pi}{6} \text { , } \frac{5\pi}{6}$

note: I have worked out the solutions of $0 \leq \theta \leq 2\pi$

**daunder** · April 26th 2009, 06:01 AM

Thanks for the quick reply

**Moo** · April 26th 2009, 06:01 AM

Originally Posted by daunder

$2sin\theta cos\theta-cos\theta=0$

This is what I did

First I took $cos\theta$ to other side resulting in $2sin\theta cos\theta=cos\theta$
Then I divided by $cos\theta$ resulting in $2sin\theta =1$ then I divided by 2 $sin\theta = \frac{1}{2}$
I am sure it's incorrect.. Can someone please help me

You can't divide by $\cos(\theta)$ as it can be equal to 0.

Here is how to do.

factorise by cos(theta) :

$\cos \theta (2 \sin \theta-1)=0$

the product ab=0 if and only if a=0 or b=0

----> $\cos \theta=0$ or $\sin \theta=\frac 12$

For the first one... there are infinitely many solutions !
we know that $\cos \frac \pi 2=0$
but since cos(-x)=cos(x), we know that $\cos \left(-\frac \pi 2\right)=0$
moreover, cos is a 2pi-periodic function.
Hence $\boxed{\cos \left(\frac \pi 2 +2k\pi\right)=\cos \left(-\frac \pi 2 +2k\pi\right)=0}$
these give all the possible solutions for cos(theta)=0

(in general, if a is such that cos(a)=b, then the solutions to cos(x)=b are in the form a+2kpi or -a+2kpi, for any integer k)

For the second one, there are infinitely many solutions too.
we know that $\sin\left(\frac \pi 3\right)=0$
but we know that $\sin(\pi-x)=\sin(x)$ . Thus $\sin\left(-\frac \pi 3+\pi\right)=0$
again, sin is a 2pi-periodic function.
Hence $\sin\left(2k\pi\right)=\sin\left(-\frac \pi 3+\pi+2k\pi\right)=0$
this can be rewritten :
$\boxed{\sin\left(2k\pi\right)=\sin\left(-\frac \pi 3+(2k+1)\pi\right)=0}$

Finally, the solutions are :
$S=\left\{\tfrac{\pi}{2}+2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{-\tfrac{\pi}{2}+2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{2k \pi \big| k \in \mathbb{Z}\right\} \cup \left\{-\tfrac \pi 3+(2k+1)\pi \big| k \in \mathbb{Z}\right\}$

does it look good to you ?

Edit : woops, a bit late

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