Question

2 cos x + 1 is equal to sin x solve the equation,
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Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\:2cosx + 1 = sinx$

On squaring both sides, we get

$\rm :\longmapsto\: {(2cosx + 1)}^{2} = {sin}^{2}x$

$\rm :\longmapsto\: {4cos}^{2}x + 1 + 4cosx = 1 - {cos}^{2}x$

$\rm :\longmapsto\: {5cos}^{2}x + 4cosx = 0$

$\rm :\longmapsto\:cosx(5cosx + 4) = 0$

$\rm \implies\:cosx = 0 \: \: \: or \: \: \: cosx = - \: \dfrac{4}{5}$

Consider, Case - 1

$\red{\rm :\longmapsto\:cosx = 0}$

$\rm \implies\:\boxed{ \tt{ \: x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z \: }}$

Consider, Case - 2

$\red{\rm :\longmapsto\:cosx = - \dfrac{4}{5}}$

Let assume that

$\red{\rm :\longmapsto\:cosy = \dfrac{4}{5} \: so \: that \: y \: = \: {tan}^{ - 1} \dfrac{3}{4} }$

So,

$\rm :\longmapsto\:cosx = - cosy$

$\rm :\longmapsto\:cosx = cos(\pi - y)$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\: x = 2n\pi \pm \: (\pi - y)\: \forall \: n \in \: Z$

$\bf \implies\:\: x = 2n\pi \pm \: \pi \mp \: y\: \forall \: n \in \: Z$

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Additional Information :-

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