Solve $4\tan x=5\sin x$ from $x$ is $0$ to $2\pi$.

Question

I multiplied by $\cos^2$ and divided by sin to get $\cos(5\cos-4)$ apparently this is wrong and I am supposed to get $\sin= 0$ not $\cos =0$ but I don't understand at all why.

Wham I wrong and why are they correct? Thank you.

Answer 1

Hint:

$$4\frac{\sin x}{\cos x}=5\sin x$$

has the obvious solutions

$$\sin x=0$$ and

$$\cos x=\frac45.$$

Answer 2

$$4\tan x = 5\sin x$$

$\sin x = 0$ and $\tan x = 0$ are immediate solutions, resulting in $x = 0$.

For the other solutions:

$$\frac{\tan x}{\sin x} = \frac{5}{4}$$

$$\frac{\frac{\sin x}{\cos x}}{\sin x} = \frac{5}{4}$$

$$\frac{1}{\cos x} = \frac{5}{4}$$

$$\cos x = \frac{4}{5}$$

From here, you can easily solve for $x$.

Answer 3

Hint: $$4\frac{\sin x}{\cos x}=5\sin x$$ $$4\frac{\sin x}{\cos x}-5\sin x=0$$ $$\frac{4\sin x-5\sin x\cos x}{\cos x}=0$$ A fraction is equal to zero when numerator is equal to zero. Factoring numerator: $$\sin x(4-5\cos x)=0$$ Hope you can finish this up