Respuesta :
You can use the identity [tex] \cos(2x) = \cos^2(x)-\sin^2(x) [/tex] to write the equation as
[tex] 2(\cos^2(x)-\sin^2(x))+ 9\sin(x) = 6 [/tex]
Now, from the fundamental equation of trigonometry, deduce an expression for [tex] \cos^2(x) [/tex] in terms of [tex] \sin^2(x) [/tex]:
[tex] \cos^2(x)+\sin^2(x) = 1 \implies \cos^2(x) = 1-\sin^2(x) [/tex]
The equation becomes
[tex] 2(1-\sin^2(x)-\sin^2(x))+ 9\sin(x) = 6 \iff 2(1-2\sin^2(x))+ 9\sin(x) = 6 [/tex]
Simplify the left hand side and move all terms to left hand side:
[tex] 2-4\sin^2(x)+ 9\sin(x) -6 = 0 \iff -4\sin^2(x) + 9\sin(x) - 4 = 0 [/tex]
Now, if you let [tex] t = \sin(x) [/tex], this equation becomes a quadratic equation:
[tex] -4t^2 + 9t - 4 = 0 [/tex]
The two solutions of this equations are
[tex] t = \cfrac{9}{8} - \cfrac{\sqrt{17}}{8},\quad t = \cfrac{9}{8} + \cfrac{\sqrt{17}}{8} [/tex]
We must be careful, because we have to remember that [tex] t [/tex] was actually [tex] \sin(x) [/tex]. This means that [tex] t [/tex] can only assume values between -1 and 1. The second solution exceeds 1, so we reject it. So, we have
[tex] t = \cfrac{9}{8} - \cfrac{\sqrt{17}}{8} \implies \sin(x) = \cfrac{9}{8} - \cfrac{\sqrt{17}}{8} \implies x = \arcsin\left(\cfrac{9}{8} - \cfrac{\sqrt{17}}{8}\right) [/tex]
