Question

Solve the following trigonometric equation for 0 ≤ A ≤ 2π : 1+cos A=sin^2A.

Answer 1

\[\sin ^{2}A\]

Answer 2

the standard formula says \[\sin ^{2} A+\cos ^{2} A=1\] so sin ^{2} A=1-cos ^{2} A
substituting this in the equation u get 1+cos A= 1-cos ^{2} A
=> cos ^{2} A+cos A=0
=> cos A ( cos A+1)=0
=> either cos A = 0 => A = 0 as cos 0 = 0
Or cos A+1=0=>cos A = -1 => A = pi degrees as Cos pi = -1 (it falls in second quadrant where all trigonometric functions are negative other than sine)

so A is either 0 or pi

Answer 3

thanks :D