Question

Number of values of x in [0, pi] satisfying the equation 2cos2x = 1 - cos 3x is equal to
(A) 2
(B) 3
(C) 4
(D) 5​

Answer 1

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

Given Trigonometric equation is

$\red{\rm :\longmapsto\:2cos2x = 1 - cos3x \: }$

can be rewritten as

$\rm :\longmapsto\:cos3x + 2cos2x - 1 = 0$

We know,

$\boxed{ \tt{ \: cos3x = {4cos}^{3}x - 3cosx \: }}$

and

$\boxed{ \tt{ \: cos2x = {2cos}^{2}x - 1 \: }}$

So, on substituting these Identities, we get

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + 2( {2cos}^{2}x - 1) - 1 = 0$

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + {4cos}^{2}x - 2 - 1 = 0$

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + {4cos}^{2}x - 3 = 0$

can be regrouped as

$\rm :\longmapsto\: {4cos}^{3}x + {4cos}^{2}x - 3cosx - 3 = 0$

$\rm :\longmapsto\:4 {cos}^{2}x(cosx + 1) - 3(cosx + 1) = 0$

$\rm :\longmapsto\:(4 {cos}^{2}x - 3)(cosx + 1) = 0$

$\rm \implies\: {cos}^{2}x = \dfrac{3}{4} \: \: or \: \: cosx = - 1$

$\rm \implies\: cosx = \pm \dfrac{ \sqrt{3} }{2} \: \: or \: \: cosx = cos\pi$

$\rm \implies\: cosx = \dfrac{ \sqrt{3} }{2} \: or \: cosx = - \dfrac{ \sqrt{3} }{2} \: or \: \: x = \pi$

$\rm \implies\: cosx = cos \dfrac{ \pi }{6} \: or \: cosx = cos\dfrac{ 5\pi }{6} \: or \: \: x = \pi$

$\rm \implies\: x =\dfrac{ \pi }{6} \: or \: x = \dfrac{ 5\pi }{6} \: or \: \: x = \pi$

So, Number of solutions = 3

• Hence, Option (B) is correct

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More to know:-

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

Answer 2

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

Given Trigonometric equation is

$\red{\rm :\longmapsto\:2cos2x = 1 - cos3x \: }$

can be rewritten as

$\rm :\longmapsto\:cos3x + 2cos2x - 1 = 0$

We know,

$\boxed{ \tt{ \: cos3x = {4cos}^{3}x - 3cosx \: }}$

and

$\boxed{ \tt{ \: cos2x = {2cos}^{2}x - 1 \: }}$

So, on substituting these Identities, we get

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + 2( {2cos}^{2}x - 1) - 1 = 0$

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + 2( {2cos}^{2}x - 1) - 1 = 0$

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + {4cos}^{2}x - 2 - 1 = 0$

$\rm :\longmapsto\: {4cos}^{3}x - 3cosx + {4cos}^{2}x - 3 = 0$

can be regrouped as

$\rm :\longmapsto\: {4cos}^{3}x + {4cos}^{2}x - 3cosx - 3 = 0$

$\rm :\longmapsto\:4 {cos}^{2}x(cosx + 1) - 3(cosx + 1) = 0$

$\rm :\longmapsto\:(4 {cos}^{2}x - 3)(cosx + 1) = 0$

$\rm \implies\: {cos}^{2}x = \dfrac{3}{4} \: \: or \: \: cosx = - 1$

$\rm \implies\: cosx = \pm \dfrac{ \sqrt{3} }{2} \: \: or \: \: cosx = cos\pi$

$\rm \implies\: cosx = \dfrac{ \sqrt{3} }{2} \: or \: cosx = - \dfrac{ \sqrt{3} }{2} \: or \: \: x = \pi$

$\rm \implies\: cosx = cos \dfrac{ \pi }{6} \: or \: cosx = cos\dfrac{ 5\pi }{6} \: or \: \: x = \pi$

$\rm \implies\: x =\dfrac{ \pi }{6} \: or \: x = \dfrac{ 5\pi }{6} \: or \: \: x = \pi$

So, Number of solutions = 3

• Hence, Option (B) is correct

▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know:-

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

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