Respuesta :
Rewrite the equation by using the definition of the tangent function:
[tex] \cfrac{5\sin(x)}{\cos(x)} + 2\cos(x) = 0[/tex]
Since there is a cosine at the denominator, it can't be zero. So, we know that
[tex] \cos(x) \neq 0 \implies x \notin \left\{\cfrac{\pi}{2},\ \cfrac{3\pi}{2}\right\} [/tex]
With this restriciton, let's multiply the whole equation by cos(x):
[tex] 5\sin(x) + 2\cos^2(x) = 0[/tex]
We don't like having sine and cosine in the same equation. We can fix this by using the fundamental identity of trigonometry to express cos^2 in terms of sin^2:
[tex] \cos^2(x)+\sin^2(x)=1 \implies \cos^2(x) = 1-\sin^2(x) [/tex]
The equation becomes
[tex] 5\sin(x) + 2(1-\sin^2(x)) = 0[/tex]
which you can rearrange as
[tex] 5\sin(x) + 2-2\sin^2(x) = 0 \iff 2\sin^2(x)-5\sin(x)-2 = 0[/tex]
This is a quadratic equation is sin(x): if you let t=sin(x) the equation becomes
[tex] 2t^2-5t-2 = 0[/tex]
which yields solutions
[tex] t = \cfrac{5}{4} - \cfrac{\sqrt{41}}{4},\quad t = \cfrac{5}{4} + \cfrac{\sqrt{41}}{4}[/tex]
Now, remember that t is sin(x), and as such it can't exceed the interval [-1,1]. Since the second solution is greater than 1, we can only accept the first solution. So, we have
[tex] t = \sin(x) = \cfrac{5}{4} - \cfrac{\sqrt{41}}{4} \implies x = \arcsin\left(\cfrac{5}{4} - \cfrac{\sqrt{41}}{4}\right) [/tex]
