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Dashdorm24 · Jun 6, 2021

Any ideas?

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

Life'sHard · Jun 6, 2021

(CosΘ - 3sinΘ) (cosΘ - 2sinΘ)=0

jimmysmith560 · Jun 6, 2021

Would it result in those values? Then we convert to radians?

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

CM_Tutor · Jun 6, 2021

$\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:

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

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

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

The second solution of the equation is:

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

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

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

And yes, @jimmysmith560, these answers in radians correspond to your answers in degrees.

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Dashdorm24

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