Question

find to the nearest degree all values of theta in the interval 0 x 360 that satisfy 3 cos 2 theta+sin theta-1=0

Answer 1

Question #28030

find to the nearest degree all values of theta in the interval 0 x 360 that satisfy 3 cos 2 theta + sin theta - 1 = 0

Solution. Using the trigonometric formulas $\cos^2\alpha + \sin^2\alpha = 1$ and $\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha$ , we can simplify:

$$3 \cos (2 \theta) + \sin \theta - 1 = 0$$ $$6 \sin^ {2} \theta - \sin \theta - 2 = 0$$

In other words we shall solve the quadratic equation.

$$D = 1 + 4 \cdot 6 \cdot 2 = 49.$$

Thus,

$$\sin \theta = \frac {1 + 7}{12} = \frac {2}{3} \text{ and } \sin \theta = \frac {1 - 7}{12} = - \frac {1}{2}.$$ $$\theta = (- 1) ^ {k} \arcsin (\theta) + 180 ^ {o} k, k \in \mathbb {Z}.$$

Since $0 \leq \theta \leq 360$ , then $\theta = 42^0, 138^0, 210^0, 330^0$ .

Answer. $42^{0}, 138^{0}, 210^{0}, 330^{0}$ .