Question

sin x + cos x = 1 find general solutions​

Answer 1

Step-by-step explanation:

sin x+cos x=1

√2[sin x/2+cos x/2] =1

sinx. cos45°+cosx.sin45°=1/√2

sin(x+pi/4)=sin(n pie+(-1)^n pie/4)

hence (x+pi/4)= n pie +(-1)^n pie/4

x= n pie +(-1) pie^n pie/4-pi/4

Answer 2

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

Given that

$\rm :\longmapsto\:sinx + cosx = 1$

$\red{ \sf \: On \: dividing \: both \: sides \: by \: \sqrt{ {1}^{2} + {1}^{2} } = \sqrt{2}, \: we \: get}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{2} }sinx + \dfrac{1}{ \sqrt{2} }cosx= \dfrac{1}{ \sqrt{2} }$

$\rm :\longmapsto\:sin\dfrac{\pi}{4} sinx + cos\dfrac{\pi}{4}cosx = cos\dfrac{\pi}{4}$

$\: \: \: \: \: \: \: \: \: \red{\: \bigg\{ \because \sf \: cos\dfrac{\pi}{4} = sin\dfrac{\pi}{4} = \dfrac{1}{ \sqrt{2} } \bigg\}}$

$\rm :\longmapsto\:cos\bigg(x - \dfrac{\pi}{4} \bigg) = cos\dfrac{\pi}{4}$

$\: \: \: \red{ \: \bigg\{ \sf\: \because \: \: cos(x - y) = cosxcosy + sinxsiny \: \bigg \}}$

$\rm :\longmapsto\:x - \dfrac{\pi}{4} = 2n\pi \: \pm \: \dfrac{\pi}{4}$

$\red{ \: \bigg\{ \sf\: \because \: \: cosx = cosy \implies \: x = 2n\pi \pm \: y \: where \: n \in \:Z \: \bigg \}}$

$\rm :\implies\:x = 2n\pi \: \: or \: x = 2n\pi + \dfrac{\pi}{2} \: where \: n \in \: Z$

Additional Information :-

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

$\: \: \rm \: \: \: \: \: \: \: \: \: \: \: \bull \: \: \: \: \: \: where \: n \in \: Z$