[tex]x = 90°, 16.3°[/tex]
Step-by-step explanation:
We are going to use the identity
[tex]\cos x =\sqrt{1-\sin^2x}[/tex]
and substitute this into our expression so we can write
[tex]4\sin x + 3\sqrt{1-\sin^2x} = 4[/tex]
[tex]\Rightarrow 3\sqrt{1-\sin^2x} = 4(1 - \sin x)[/tex]
Take the square of the equation above to get
[tex]9(1 - \sin^2x) = 16(1 - \sin x)^2[/tex]
[tex]\:\:\:\:\:= 16(1 - 2\sin x + \sin^2x)[/tex]
Rearranging the terms, we then get
[tex]25\sin^2x - 32\sin x + 7 = 0[/tex]
If we let [tex]u = \sin x,[/tex] the above equation becomes
[tex]25u^2 - 32u + 7= 0[/tex]
This looks like a quadratic equation whose roots are
[tex]u = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]\;\;\;\;= \dfrac{32 \pm \sqrt{32^2 - 4(25)(7)}}{50}[/tex]
[tex]\;\;\;\;=\dfrac{32 \pm 18}{50} = 1, 0.28[/tex]
We can then write
[tex]\sin x = 1 \:\text{and}\;0.28[/tex]
Solving for x, we finally get
[tex]x = 90°\:\text{and}\:16.3°[/tex]
