The equation in the title should be not . I want to solve for . After a few steps I get or . From equation , at first I assume that and I found that is one solution. From equation , in what way can I proceed for the case ?
7 sin(2x) - cos(2x) = 1
$14\sin{x}\cos{x} - (\cos^2{x} - \sin^2{x}) = \cos^2{x} + \sin^2{x}$
$14\sin{x}\cos{x} - 2\cos^2{x} = 0$
$\cos{x}(14\sin{x} - 2\cos{x}) = 0$
$\cos{x} = 0$
$\tan{x} = \dfrac{1}{7}$
for $0 \le x < \pi$ ...
$x = \dfrac{\pi}{2}$
$x = \arctan\left(\dfrac{1}{7}\right)$
From $\cos{x}(14\sin{x} - 2\cos{x}) = 0$, we get $\cos{x}=0$ or $14\sin{x}-2\cos{x}=0$. From equation $14\sin{x}-2\cos{x}=0$, by assuming that $sin{x}\neq{0}$, we get the solution $x=\tan^{-1}(1/7)$. However, for the equation $14\sin{x}-2\cos{x}=0$, we have not yet rule out the possibility that $sin{x}=0$ since we previously assume that $sin{x}\neq{0}$. How can we show that x for which $\sin{x}=0$ is not a solution of $14\sin{x}-2\cos{x}=0$ and hence not a solution of the original equation?
Why do you care if $\sin x=0$ at all? You should be caring about $\cos x \neq 0$:
$7\sin x = \cos x$
$7\dfrac{\sin x}{\cos x} = 1$
$\tan x = \dfrac{1}{7}$
You are dividing by $\cos x$. Fortunately, if $\cos x=0$, you already have a solution, so you can assume that $\cos x \neq 0$ and then solve.