cosx + sin 2x =0; 0 less than or equals to x less than 2pi

Answer 1

We may rewrite the equation as cosx+2(sinx)cosx=0 which is the same as (cosx)(1+2sinx)=0. Hence, either cosx=0 or sinx=-1/2. But since x is in the interval [0,2π) we obtain that either x=π/2 or 3π/2 (from cosx=0) or x=7π/6 or 11π/6 (from sinx=-1/2).