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  4. The most general solution of sin θ=-(1/2) and tan θ=(1/√3) are

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Q. The most general solution of $\sin \theta=-\frac{1}{2}$ and $\tan\,\theta=\frac{1}{\sqrt{3}}$ are

Trigonometric Functions

Solution:

$\sin\,\theta=-\frac{1}{2}$ and $\tan\,\theta=\frac{1}{\sqrt{3}}$
$\therefore \,\theta$ lies in the IIIrd quadrant.
$\therefore \, \, \,\sin\,\theta=\sin\,\left(\pi+\frac{\pi}{6}\right)=\sin\,\frac{7\pi}{6}$
$\, \, \, \, \, \, \cos\,\theta=\cos\,\left(\pi+\frac{\pi}{6}\right)=\cos\,\frac{7\pi}{6}$
$\, \, \, \, \, \, \tan\,\theta=\tan\,\frac{7\,\pi}{6}\,\therefore \,\theta=2n\pi+\frac{7\pi}{6}$