Show that $4 - \cos{2\theta} + 5\sin{\theta} = 2\sin^2{\theta} + 5\sin{\theta} +3$. Hence, so

Question

Show that $4 - \cos{2\theta} + 5\sin{\theta} = 2\sin^2{\theta} + 5\sin{\theta} +3$. Hence, solve the equation $4 - \cos{2\theta} + 5\sin{\theta} = 0$ for $0 \leq \theta \leq 2\pi$.

Accepted Answer

(a) $4-\cos2\theta+5\sin\theta=2\sin^2\theta+5\sin\theta+3$ (b) $\theta=\dfrac{3\pi}2$

Trigonometry - Trigonometric Equations

Trigonometry - Trigonometric Equations

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