Originally Posted by

**Peleus** Hi all,

To start with - the problem.

$\displaystyle 3\sin^2\theta + 7\cos\theta = 5$ solve for $\displaystyle 0 <= \theta <= 2\pi $

The working I've done so far.

1. $\displaystyle 3(1 - \cos^2\theta) + 7\cos\theta = 5$

2. $\displaystyle 3 - 3\cos^2\theta + 7\cos\theta = 5$

3. $\displaystyle \cos\theta (7 - 3\cos\theta) = 2 $

4. $\displaystyle \cos\theta = 2$ or $\displaystyle (7 - 3\cos\theta) = 2$

5. $\displaystyle 7 - 3\cos\theta = 2 $

6. $\displaystyle -3\cos\theta = -5$

7. $\displaystyle \cos\theta = \frac{3}{5}$

Therefore, $\displaystyle \theta = 53.13 deg$ or $\displaystyle \theta = 306.87 deg$

Which means that $\displaystyle \theta = \frac{53.13\pi}{180}$ or $\displaystyle \theta = \frac{306.87\pi}{180}$

I'm just not sure if step 4 is correct. Can I simply ignore one of the options because $\displaystyle \cos\theta = 2$ is invalid?

Cheers.