Question

sinx+cosx=1+sinxcosx solve for x

Answer 1

by squaring both sides we have
(sinx)^2+(cosx)^2+2 cosxsinx = 1+(sinx)^2(cosx)^2+2cosxsinx
1+2cosxsinx=1+(sinx)^2(cosx)^2+2cosxsinx
(sinx)^2(cosx)^2=0
sinx cosx=0
x=0,90°

Answer 2

Answer: 0° or 90°

Step-by-step explanation:

Given that : $\sin x+\cos x=1+\sin x\cos x$

Squaring both side we get

$(\sin x+\cos x)^{2} =(1+\sin x\cos x)^{2} \\\\\Rightarrow(\sin x)^{2} +(\cosx)^{2} +2\sin x\cos x=1+(\sin x\cos x)^{2} +2\sin x\cos x\\\\\Rightarrow \sin^2x +\cos^2x =1+ \sin^2x\cos^2x\\\\\Rightarrow1=1+\sin^2x\cos^x\\\\\Rightarrow \sin^2x\cos^2x=0\\\\\Rightarrow \sin x\cos x=0$

Therefore either $\sin x=0$ or $\cos x=0$

i.e. either $x=0^\circ$ or $x=90^\circ$

Hence, the value of x is 0° or 90°