Answer:
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:
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#
