We have: #2 sin^(2)(x) + 3 sin(x) = 4#; #[0, 360^(circ))#
If you take a look at the equation, you may notice that it is almost in the form of a quadratic equation.
Let's subtract #4# from both sides of the equation:
#Rightarrow 2 sin^(2)(x) + 3 sin(x) - 4 = 0#
We can consider #sin(x)# to be a variable of its own, so let's apply the quadratic formula:
#Rightarrow sin(x) = frac(- 3 pm sqrt(3^(2) - 4(2)(- 4)))(2(2))#
#Rightarrow sin(x) = frac(- 3 pm sqrt(9 + 32))(4)#
#Rightarrow sin(x) = frac(- 3 pm sqrt(41))(4)#
#Rightarrow sin(x) = frac(- 3 pm 6.40)(4)#
#Rightarrow sin(x) = - 2.35, 0.85#
However, the range of #sin(x)# is #[- 1, 1]#.
So #sin(x) = 0.85#.
Let's set the reference angle as #sin(x) = 0.85 Rightarrow x = 58^(circ)#:
#Rightarrow sin(x) = 0.85#
#Rightarrow x = 58^(circ), 180^(circ) - 58^(circ)#
#therefore x = 58^(circ), 122^(circ)#
Therefore, the solutions to the equation are #x = 58^(circ)# and #x = 122^(circ)#.
