Let's simplify the problem down a little:
⇒ set 'sin(θ)' as x
⇒ the new equation looks like: [tex]2x^2+5x-3=0[/tex]
Now let's solve this equation as a quadratic:
[tex]2x^2+5x-3=0\\(2x-1)(x+3)=0\\\\2x-1=0\\x = \frac{1}{2} \\\\x+3 = 0\\x=-3[/tex]
Now set the x value of '1/2', to sin(θ)
- it can't be -3, because trig functions can't equal a negative number
sin(θ) = 1/2
θ = π/6, which is between the range given
Answer: π/6
Hope that helps!
