Question

Solve: sinx+ cosx = 0 in [-pi/2,pi/2]

Answer 1

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

Given that,

$\red{\rm :\longmapsto\:x \: \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi}{2} \bigg]}$

and

$\rm :\longmapsto\:sinx + cosx = 0$

can be rewritten as

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

$\rm :\longmapsto\:\dfrac{sinx}{cosx} \: = \: - \: 1$

$\rm :\longmapsto\:tanx \: = \: - \: 1$

$\rm :\longmapsto\:tanx \: = \: - \: tan\bigg[\dfrac{\pi}{4} \bigg]$

can be rewritten as

$\rm :\longmapsto\:tanx \: = \: tan\bigg[ - \: \dfrac{\pi}{4} \bigg]$

$\bf\implies \:\boxed{ \bf{ \: x \: = \: - \: \dfrac{\pi}{4} \: }}$

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Additional Information :-

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