Question

How do you solve `2sin^2x+3sinx+1=0` and find all solutions in the interval `[0,2pi)`?

Answer 1

`x= (7pi)/6 or (11pi)/6 or (3pi)/2`

Explanation:

We may identify this as a quadratic equation in `sinx` and factorise it as a trinomial to obtain the solutions as

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

`thereforesinx=-1/2 or sinx=-1`

`therefore x= (7pi)/6 or (11pi)/6 or (3pi)/2`