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

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

Explanation:

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

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.