Question

How do you solve `2cos^2x = sinx + 1` from `[0,2pi]`?

Answer 1

x= `pi/6, (5pi)/6`,`(3pi)/2`

Explanation:

Re write it as `2(1- sin^2x)= sinx +1`,

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

(2sinx -1)(sinx +1)=0

sin x= `1/2` , -1

sin x =`1/2` would have x= `pi/6, (5pi)/6` in the prescribed interval

sin x= -1 would have x= `(3pi)/2`