Question

cos^2 2x + cos^2 x = 1 then general solution?

Answer 1

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

Given Trigonometric equation is

$\rm \: {cos}^{2}2x + {cos}^{2}x = 1$

On multiply both sides by 2, we get

$\rm \: 2{cos}^{2}2x + 2{cos}^{2}x = 2$

We know,

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

So, using this identity, we get

$\rm \: {2cos}^{2}2x + 1 + cos2x = 2$

$\rm \: {2cos}^{2}2x + 1 + cos2x - 2 = 0$

$\rm \: {2cos}^{2}2x + cos2x - 1= 0$

On splitting the middle terms, we get

$\rm \: {2cos}^{2}2x + 2cos2x - cos2x - 1= 0$

$\rm \: 2cos2x(cos2x + 1) - 1(cos2x + 1) = 0$

$\rm \: (cos2x + 1)(2cos2x - 1) = 0$

$\rm\implies \:cos2x = - 1 \: \: or \: \: cos2x = \frac{1}{2}$

So, Consider when

$\rm \: cos2x = - 1$

can be rewritten as

$\rm \: cos2x = cos\pi$

We know,

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

So, using this result, we get

$\rm \: 2x = 2n\pi \pm \: \pi \: \: \: \: \: \forall \: n \in \: Z$

$\rm\implies \:\rm \: x = n\pi \pm \: \frac{\pi}{2} \: \: \: \: \: \forall \: n \in \: Z$

Now, Consider when

$\rm \: cos2x = \frac{1}{2}$

can be rewritten as

$\rm \: cos2x = cos\frac{\pi}{3}$

$\rm \: 2x = 2m\pi \pm \: \frac{\pi}{3} \: \: \: \: \forall \: \: m \: \in \: Z$

$\rm\implies \:\rm \: x = m\pi \pm \: \frac{\pi}{6} \: \: \: \: \forall \: \: m \: \in \: Z$

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Additional Information

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