Question

2sin^2x - 5sinx - 3=0
Part 1: Rewrite the equation by substituting the expression u in for sin x.

Part 2: Factor the quadratic expression. Rewrite the equation with factors instead of the original polynomial.

Part 3: Use the zero product property to solve the quadratic equation.

Part 4: Rewrite your solutions to part 3 by replacing u with sin x.

Part 5: Solve the remaining equations for x, giving all solutions to the equation.

Answer 1

well this is an equation that can be rewritten as a quadratic.

making the required substitution gives

\[2u^2 -5u - 3 = 0\]

so now you are solving the quadratic

Answer 2

so next you need to factor the quadratic...

Answer 3

how would i solve that?

Answer 4

well here is my method

for a quadratic \[ax^2 + bx + c = 0\]

multiply a and c

so in your question you need to multiply 2 and -3

then find the factors of the answer that add to -5... the larger factor is negative.

Any thoughts..?

Answer 5

So for the factors do I just pick any numbers that add up to -5, like -10+5?

Answer 6

nope if you multiply 2 and -3 you get -6

you need to find the factors of -6 that add to -5... the larger is negative

Answer 7

oh so like -6+1

Answer 8

thats it,,, next part to factoring

write it as (ax + factor 1)(ax + factor 1)
------------------------- = 0
a

so you get

(2u -6)(2u + 1)
--------------- = 0
2

Answer 9

then remove the common factor in the 1st binomial

2(u - 3)(2u + 1)
---------------- = 0
2

so the factored form is

(u -3)(2u + 1) = 0

now you need to solve it

Answer 10

would it be 2u^2-5u-3=0 ?

Answer 11

now all that has been done is to write the original equation in factored form. As required by Part 2.

Part 3

let

u - 3 = 0

2u + 1 = 0

solve both equations for u....

Answer 12

so u=3 and u= -1/2

Answer 13

I still need help with part 5