Question

Solve 2 sin theta cos theta = 2 - sin theta + 4 cos theta

Answer 1

$\large\underline{\sf{Solution-}}$

Given Trigonometric equation is

$\red{\rm :\longmapsto\:2sin\theta cos\theta = 2 - sin\theta + 4cos\theta }$

can be rewritten as

${\rm :\longmapsto\:2sin\theta cos\theta - 2 + sin\theta - 4cos\theta = 0}$

can be further re - arranged as

${\rm :\longmapsto\:2sin\theta cos\theta + sin\theta - 2 - 4cos\theta = 0}$

$\rm :\longmapsto\:sin\theta (2cos\theta + 1) - 2(1 + 2cos\theta ) = 0$

$\rm :\longmapsto\:sin\theta (2cos\theta + 1) - 2(2cos\theta + 1 ) = 0$

$\rm :\longmapsto\:(sin\theta - 2) (2cos\theta + 1)= 0$

$\bf\implies \:sin\theta = 2 \: \{rejected \: as \: - 1 \leqslant sin\theta \leqslant 1 \}$

So,

$\rm :\longmapsto\:2cos\theta + 1 = 0$

$\rm :\longmapsto\:2cos\theta = - \: 1$

$\rm :\longmapsto\:cos\theta \: = - \: \dfrac{1}{2}$

$\rm :\longmapsto\:cos\theta \: = cos \: \dfrac{2\pi}{3}$

We know,

$\purple{\boxed{ \rm{ cosx = cosy\bf\implies \:x = 2n\pi \pm \: y \: \forall \: n \in \: Z}}}$

So, using this identity, we get

$\bf\implies \:\theta = 2n\pi \: \pm \: \dfrac{2\pi}{3} \: \forall \: n \in \: Z$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2} \: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z \\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y \: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$