Math, asked by krisdattani05, 1 month ago

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

Answers

Answered by mathdude500
5

\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}

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