Question

cos²x +sinx=1 <-- find all solutions of the equation in the interval [0, 2pi)

Answer 1

\[\cos ^{2}x+sinx=1-\sin ^{2}x+sinx=1\] so \[\sin ^{2}x=sinx\], which means x=0 or x=pi.

Answer 2

wait, real quick, if sin2x= radical2/2 , then how do you symplify that?

Answer 3

\[\sin 2x =\sqrt{2}\div2\]

Answer 4

That indicates that 2x is the angle that gives sin value of \[\sqrt{2}/2\]. That would indicate a value of x of \[\pi /8+2k \pi\]or\[3\pi/8+2k \pi\]