Solution
$$\sin5\theta=\cos(50^\circ-3\theta)$$
$$\cos\left(90^{\circ\ }-5\theta\ \right)=\cos\left(50^{\circ\ }-3\theta\ \right)$$
$$90^{\circ\ }-5\theta\ =50^{\circ\ }-3\theta\ $$
$$\theta=20^{\circ\ }\ $$
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