Question

solve sinθ = √3(1-cosθ) where 0<=θ<=360

Answer 1

${ \boxed{\boxed{\begin{array}{cc} \bf sin \theta = \sqrt{3} (1 - cos\theta) \\ \\ \bf \implies \: sin\theta = \sqrt{3} - \sqrt{3} cos\theta \\ \\ \bf \implies \: \sqrt{3} cos\theta + sin\theta = \sqrt{3 } \\ \\ \pink{\bf \: now \: divided \: both \: side \: by \: 2} \\ \\\bf \implies \: \frac{ \sqrt{3} }{2} cos\theta + \frac{1}{2}sin\theta = \frac{ \sqrt{3} }{2} \\ \\ \bf \implies \: cos {30}^{ \circ} \: cos\theta + sin {30}^{ \circ} \: sin \theta = cos {30}^{ \circ} \\ \\ \small{\pink{ {\boxed{\begin{array}{cc} \bf we \: know \: that \: \\ \\ \bf \: cos(x - y) = cos x \: cosy + sinx \: siny\end{array}}}} }\: \\ \\ \bf \implies \: cos(\theta - {30}^{ \circ} ) = cos {30}^{ \circ} \\ \\ \pink{ {\boxed{\begin{array}{cc} \bf \: we \: know \: that \\ \\ \bf \: if \: \: cos\theta = cos \alpha \\ \\ \bf \: then \: \: \theta =2 n\pi \pm \: \alpha \: \: \: where \: n \in \: Z \end{array}}}} \\ \\ \bf \implies \: \theta - {30}^{ \circ} = 2 n\pi \pm{30}^{ \circ} \\ \\ \bf \implies \: \theta =2 n\pi \pm {30}^{ \circ} + {30}^{ \circ} \end{array}}}}$

Now,

Taking (+ve)

$\bf \: \theta = 2\pi + 30 + 30 \\ = \bf2n\pi + 60$

Taking (-ve)

$\bf\theta = 2n\pi - 30 + 30 \\ = \bf 2n\pi \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Again given,

$\bf0 \leqslant \theta \leqslant 360$

${ {\boxed{\begin{array}{c |c| c} \underline{\bf \: value \bf \: of \: n}& \underline{\bf\theta = 2n\pi + 60}& \underline{\theta = 2n\pi }\\ \\ \bf \: n = 0&\theta = 60&\theta = 0 \\ \\ \bf \: n = 1&\theta = 420&\theta = 360 \\ \\ \bf \: n = - 1&\theta = - 300&\theta = - 360 \\ \\ \bf ...&...&...\end{array}}}}$

According to the range,

$\theta \: = {0}^{ \circ} , {60}^{ \circ}, {360}^{ \circ}$