# Math Help - 2 cos \theta - 4 sin^2 \theta = 0

1. ## 2 cos \theta - 4 sin^2 \theta = 0

Really struggling with trig functions, can anyone give me a step by step how to find angles between 0 and 360 please

$2 cos \theta - 4 sin^2 \theta = 0$

Thanks,

Kris

2. ## Re: 2 cos \theta - 4 sin^2 \theta = 0

Originally Posted by Krislton
Really struggling with trig functions, can anyone give me a step by step how to find angles between 0 and 360 please

$2 cos \theta - 4 sin^2 \theta = 0$

Thanks,

Kris
\displaystyle \begin{align*} 2\cos{(\theta)} - 4\sin^2{(\theta)} &= 0 \\ 2\cos{(\theta)} - 4\left[ 1 - \cos^2{(\theta)} \right] &= 0 \\ 2\cos{(\theta)} - 4 + 4\cos^2{(\theta)} &= 0 \\ 4x^2 + 2x - 4 &= 0 \textrm{ if we let }x = \cos{(\theta)} \end{align*}

Now solve for x using whatever method you prefer to solve the quadratic equation, and use this to find the values of theta.

3. ## Re: 2 cos \theta - 4 sin^2 \theta = 0

First, you need to know some trig identities- like sin^2x+ cos^2x= 1. In this particular problem, that lets you replace " $sin^2\theta$" with [tex]1- cos^2(\theta)[/itex] so the problem becomes [tex]2cos(\theta)- 4(1- cos^2(\theta))= 4cos^2(\theta)+ 2 cos(\theta)- 4= 0[/itex].

Do you recognize that as a quadratic equation? If you replace " $cos(\theta)$" with "y" you have [tex]4y^2+ 2y- 4= 0[/quote]. You can use the quadratic formula to solve that, then use a calculator to solve $cos(\theta)= y$ for one value of $\theta$. I don't know if you have learned the graph of cosine or use the "unit circle" but you should know that there are two values of $\theta$, between 0 and 360, such that $cos(\theta)= y$. (That is, if y is between -1 and 1. If not, there is no solution.) There are either 0, or 2, or 4 solutions to this equation between 0 and 360.

4. ## Re: 2 cos \theta - 4 sin^2 \theta = 0

Originally Posted by HallsofIvy
First, you need to know some trig identities- like sin^2x+ cos^2x= 1. In this particular problem, that lets you replace " $sin^2\theta$" with [tex]1- cos^2(\theta)[/itex] so the problem becomes [tex]2cos(\theta)- 4(1- cos^2(\theta))= 4cos^2(\theta)+ 2 cos(\theta)- 4= 0[/itex].

Do you recognize that as a quadratic equation? If you replace " $cos(\theta)$" with "y" you have [tex]4y^2+ 2y- 4= 0. You can use the quadratic formula to solve that, then use a calculator to solve $cos(\theta)= y$ for one value of $\theta$. I don't know if you have learned the graph of cosine or use the "unit circle" but you should know that there are two values of $\theta$, between 0 and 360, such that $cos(\theta)= y$. (That is, if y is between -1 and 1. If not, there is no solution.) There are either 0, or 2, or 4 solutions to this equation between 0 and 360.
Yeah, I was trying to get it to a quadratic but I don't know my trig identities well enough to convert the sin to cos. I really appreciate your help, and I spend so many hours a day on math trying to understand, just sometimes I really struggle to see a link between identities. I'm sure I'll understand one day lol.