Question

help me in solving the question.

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Answer 1

$\large\underline{\sf{Given \:Question - }}$

Solve the following equation :-

$\rm :\longmapsto\: {2sin}^{2}x + 2 \sqrt{2} sinx - 3 = 0$

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

Given Trigonometric equation is

$\rm :\longmapsto\: {2sin}^{2}x + 2 \sqrt{2} sinx - 3 = 0$

On splitting the middle terms, we get

$\rm :\longmapsto\: {2sin}^{2}x + 3 \sqrt{2} sinx - \sqrt{2} sinx- 3 = 0$

can be rewritten as

$\rm :\longmapsto\: \sqrt{2} \times \sqrt{2} {sin}^{2}x + 3 \sqrt{2} sinx - \sqrt{2} sinx- 3 = 0$

$\rm :\longmapsto\: \sqrt{2}sinx( \sqrt{2}sinx + 3) - 1( \sqrt{2}sinx + 3) = 0$

$\rm :\longmapsto\: (\sqrt{2}sinx - 1)( \sqrt{2}sinx + 3) = 0$

$\bf\implies \:sinx = \dfrac{1}{ \sqrt{2} } \: \: or \: \: - \dfrac{3}{ \sqrt{2} } \: \{rejected \}$

$\bf\implies \:sinx =sin \dfrac{\pi}{4} \:$

We know,

$\boxed{ \tt{ \: sinx = siny \: \rm \implies\:y = n\pi + {( - 1)}^{n}y\: \forall \: n \in \: Z}}$

So, using this identity, we get

$\red{\bf\implies \:\boxed{ \tt{ \: x = n\pi + {( - 1)}^{n}\dfrac{\pi}{4} \: \: \forall \: n \in \: Z \: \: }}}$

Additional Information :-

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

Answer 2

Answer:

$\large\underline{\sf{Given \:Question - }}$

Solve the following equation :-

$\rm :\longmapsto\: {2sin}^{2}x + 2 \sqrt{2} sinx - 3 = 0$

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

Given Trigonometric equation is

$\rm :\longmapsto\: {2sin}^{2}x + 2 \sqrt{2} sinx - 3 = 0$

On splitting the middle terms, we get

$\rm :\longmapsto\: {2sin}^{2}x + 3 \sqrt{2} sinx - \sqrt{2} sinx- 3 = 0$

can be rewritten as

$\rm :\longmapsto\: \sqrt{2} \times \sqrt{2} {sin}^{2}x + 3 \sqrt{2} sinx - \sqrt{2} sinx- 3 = 0$

$\rm :\longmapsto\: \sqrt{2}sinx( \sqrt{2}sinx + 3) - 1( \sqrt{2}sinx + 3) = 0$

$\rm :\longmapsto\: (\sqrt{2}sinx - 1)( \sqrt{2}sinx + 3) = 0$

$\bf\implies \:sinx = \dfrac{1}{ \sqrt{2} } \: \: or \: \: - \dfrac{3}{ \sqrt{2} } \: \{rejected \}$

$\bf\implies \:sinx =sin \dfrac{\pi}{4} \:$

We know,

$\boxed{ \tt{ \: sinx = siny \: \rm \implies\:y = n\pi + {( - 1)}^{n}y\: \forall \: n \in \: Z}}$

So, using this identity, we get

$\red{\bf\implies \:\boxed{ \tt{ \: x = n\pi + {( - 1)}^{n}\dfrac{\pi}{4} \: \: \forall \: n \in \: Z \: \: }}}$

Additional Information :-

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