Given the equation cos^2 x-sinx-sin^2 x=0, find the solution that lies in the interval 0

cos^2 x-sinx-sin^2 x=0      writing cos^2x as 1 - sin^2x, we have

1-sin^x - sinx - sin^2x = 0

-2sin^x - sinx + 1  =  0        ....multiply through by -1

2sin^x + sinx - 1 = 0            .....factor

(2sinx - 1) (sinx + 1) = 0      ....set each factor to 0

2sinx - 1 = 0            sinx + 1 - 0

sinx = 1/2                sinx = -1

x = 30° , 150°  and 270°

So....x = 30°