Question

sin^2-sinx-1=0

Answer 1

\(\bf sin^2(x)-sinx-1=0\implies [{\color{brown}{ sin(x)}}]^2-{\color{brown}{ sin(x)}}-1=0\)

so solve the quadratic function to find what "sin(x)" is, and then get the \(\bf sin^{-1}\) to find what "x" is