Question

Solve on the interval [0,2pi):
2 sin^2 x-3 sin x + 1 = 0

Answer 1

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

Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations

Answer:

Step-by-step 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#