Question

Solve the equation to find general solution

$\sf \: {2sin}^{2}x - 5sinx \: cosx - {8cos}^{2}x = - 2$
​

Answer 1

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

Given Trigonometric equation is

$\rm \: {2sin}^{2}x - 5sinx \: cosx \: - \: {8cos}^{2}x = - 2$

To simplify this, Let's first check by substituting cosx = 0, whether cosx is 0 or not.

So, we get

$\rm \: {2sin}^{2}x = - 2$

$\rm \: {sin}^{2}x = - 1$

which is not possible as square number can never be negative.

So,

$\rm\implies \:cosx \: \ne \: 0$

Now, Consider again

$\rm \: {2sin}^{2}x - 5sinx \: cosx \: - \: {8cos}^{2}x = - 2$

can be rewritten as

$\rm \: \dfrac{{2sin}^{2}x - 5sinx \: cosx \: - \: {8cos}^{2}x}{ {cos}^{2}x } = - \dfrac{2}{ {cos}^{2}x}$

$\rm \: 2 {tan}^{2}x - 5tanx - 8 = - 2 {sec}^{2}x$

$\rm \: 2 {tan}^{2}x - 5tanx - 8 = - 2 (1 + {tan}^{2}x)$

$\rm \: 2 {tan}^{2}x - 5tanx - 8 = - 2 - 2 {tan}^{2}x$

$\rm \: 4{tan}^{2}x - 5tanx - 6 = 0$

$\rm \: 4{tan}^{2}x - 8tanx + 3tanx - 6 = 0$

$\rm \: 4tanx(tanx - 2) + 3(tanx - 2) = 0$

$\rm \: (tanx - 2)(4tanx + 3) = 0$

$\rm\implies \:tanx = 2 \: \: or \: \: tanx = - \dfrac{3}{4}$

Let consider

$\rm \: tanx = 2$

$\rm \: tanx = tan \alpha \: \: \: \: where \: \alpha = {tan}^{ - 1}2$

$\rm\implies \:x = n\pi + \alpha \: \: \: \: \forall \: n \in \: Z$

$\rm\implies \:x = n\pi + {tan}^{ - 1}2 \: \: \: \: \forall \: n \in \: Z$

Now, Consider

$\rm \: tanx = - \dfrac{3}{4}$

$\rm \: tanx = tan \beta \: \: \: where \: \beta = {tan}^{ - 1}\bigg( - \dfrac{3}{4}\bigg)$

$\rm\implies \:x = p\pi + \beta \: \: \: \: \: \forall \: p \in \: Z$

$\rm\implies \:x = p\pi + {tan}^{ - 1}\bigg( - \dfrac{3}{4}\bigg) \: \: \: \: \: \forall \: p \in \: Z$

$\rm\implies \:x = p\pi - {tan}^{ - 1}\bigg(\dfrac{3}{4}\bigg) \: \: \: \: \: \forall \: p \in \: Z$

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FORMULA USED

$\boxed{\tt{ \: \: tanx = tany \: \: \rm\implies \: \sf x = n\pi + y \: \: \: \: \forall \: n \in \: Z \: \: }}$

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ADDITIONAL INFORMATION

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

Answer 2

$\sf\red{Solution:-}$

• Answer in the above attachment

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@Shivam

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