Solving trig equation where $\sin (\theta + 60°) = \sin 2 \theta$

Question

I solved the equation $\sec θ + \sqrt{3} \cscθ = 4$

The question required us to use $\sinθ + \sqrt{3} \cosθ = 2 \sin 2θ$. After doing all of this I ended up with

$\sin(θ+60)=\sin(2θ)$

The range is $0< θ < 180$ and the solutions $60, 40 \text { and } 160$

I understand why its $60$ but not the rest.

Does anyone mind explaining the reasoning for this?

Thank you.

Answer 1

We know that $sin(180^o - \theta) = sin(\theta) and sin(180^o + \theta) = - sin(\theta)= sin(-\theta)$

So case-1, using $sin(\theta + 60) = sin(2\theta), \theta + 60 = 2\theta$

or$ \theta = 60$

case-2,using $sin(\theta + 60) = sin(180 - \theta - 60) = sin(2\theta)$ $$ sin(120 - \theta) = sin(2\theta)$$ $$ 120 - \theta = 2\theta$$ $$ 120 = 3\theta$$ $$ \theta = 40$$

Aliter

$sin(60+\theta) = sin(2\theta)$ $$ sin(60+\theta) - sin(2\theta) = 0$$ Applying$ sin A - sin B = 2 cos \frac{A+B}{2} sin\frac{A-B}{2}$ $$2cos\frac{60+\theta+2\theta}{2}sin\frac{60+\theta - 2\theta}{2} = 0$$ $$cos(30 + \frac{3\theta}{2})sin(30 - \frac{\theta}{2})= 0$$

Now either$$ cos(30 + \frac{3\theta}{2}) = 0$$ or$$sin(30 - \frac{\theta}{2})= 0$$ $ cos(30 + \frac{3\theta}{2}) = 0 = cos 90$ or $cos 270$

Taking 90, $30 + \frac{3\theta}{2} = 90, \frac{3\theta}{2} = 60, \theta = 40$

Taking 270, $30 + \frac{3\theta}{2} = 270, \frac{3\theta}{2} = 240, \theta = 160$

Similarly take sin 0 = sin 180 = 0, to get 60(case 1).

Thank you.