# Need help with question. (1 Viewer)

#### Dashdorm24

##### New Member
Any ideas?

Solve cos^2theta - 5cos theta * sin theta + 6sin^2theta =0 between -pi ≤ theta ≤ pi

#### Life'sHard

##### Well-Known Member
(CosΘ - 3sinΘ) (cosΘ - 2sinΘ)=0

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would it result in those values? Then we convert to radians?

$\bg_white Degrees:\:\theta =18.43494^{\circ },\:\theta =-161.56505^{\circ },\:\theta =26.56505^{\circ },\:\theta =-153.43494^{\circ }$

#### CM_Tutor

##### Moderator
Moderator
\bg_white \begin{align*} \cos^2{\theta} - 5\cos{\theta} \times \sin{\theta} + 6\sin^2{\theta} &= 0 \quad \text{between -\pi \leqslant \theta \leqslant \pi} \\ \cos^2{\theta} - 2\cos{\theta}\sin{\theta} - 3\cos{\theta}\sin{\theta} + 6\sin^2{\theta} &= 0 \\ \cos{\theta}(\cos{\theta} - 2\sin{\theta}) + 3\sin{\theta}(2\sin{\theta} - \cos{\theta}) &= 0 \\ \cos{\theta}(\cos{\theta} - 2\sin{\theta}) - 3\sin{\theta}(\cos{\theta} - 2\sin{\theta}) &= 0 \\ (\cos{\theta} - 2\sin{\theta})(\cos{\theta} - 3\sin{\theta}) &= 0 \end{align*}

The equation thus has two solutions:

$\bg_white \cos\theta - 2\sin\theta = 0 \quad \implies \quad \cfrac{\sin\theta}{\cos\theta} = \cfrac{1}{2} \quad \implies \quad \tan\theta = \cfrac{1}{2}$

This has two solutions, one in quadrant 1 and one in quadrant 3:

\bg_white \begin{align*} \theta &= \tan^{-1}{\cfrac{1}{2}} \quad \text{and} \quad \tan^{-1}{\cfrac{1}{2}} - \pi \\ &= 0.4636 \quad \text{and} \quad -2.6779 \end{align*}

The second solution of the equation is:

$\bg_white \cos\theta - 3\sin\theta = 0 \quad \implies \quad \cfrac{\sin\theta}{\cos\theta} = \cfrac{1}{3} \quad \implies \quad \tan\theta = \cfrac{1}{3}$

This also has two solutions, one in quadrant 1 and one in quadrant 3:

\bg_white \begin{align*} \theta &= \tan^{-1}{\cfrac{1}{3}} \quad \text{and} \quad \tan^{-1}{\cfrac{1}{3}} - \pi \\ &= 0.3218 \quad \text{and} \quad -2.8198 \end{align*}