Trigonometric inequality has different solutions, depending on the "path" to solution.

Question

The inequality
$$sin(x) + \sqrt3*cos(x) > 1 $$

has two solutions:

1st solution:
divide the inequality by 2

$$\frac 12*sin(x) + \frac{\sqrt{3}}{2}*cos(x) > \frac12$$
$$\frac 12$$ is $$sin(\pi/6)$$, $$\frac {\sqrt{3}} 2$$ is $$cos(\pi/6)$$ therefore
$$sin(30)*sin(x) + cos(30)*cos(x) > \frac 12$$
we can use the addition formula here (sin(a+b)) therefore
$$sin(30 + x) > \frac 12 $$ we get one solution from this:
$$\frac {\pi}{6} + 2*k*\pi< x < \frac {\pi}{2} + 2*k*\pi$$

2nd solution:

try to get just one function (turn cos into sin)
$$ sin(x) + \sqrt{3*(1-sin^2(x)} > 1 $$
$$3*(1-sin^2(x)) > 1-2*sin(x)+sin^2(x)$$
$$ 2*sin^2(x) - sin(x) - 1 < 0 $$
this quadratic inequality has two solutions:
$$ sin(x) = 1$$
$$ sin(x) = \frac{-1}{2}$$

so we get two solutions from this

$$ \frac{-\pi}{6} + 2*k*\pi < x < \frac{\pi}{2} + 2*k*\pi $$
$$ \frac{\pi}{2}+2*k*\pi < x < \frac{7*\pi}{6}+2*k*\pi $$

How is this possible?

Answer 1

That's because the equaliy $\cos x=\sqrt{1-\sin^2x}$ only holds when $\cos x\geqslant 0$.

Answer 2

The solution of $\sin x>\sin\theta$ $\;(0<\theta<\frac\pi2$) is not $\;\theta+2k\pi<x<\frac\pi2+2k\pi$ (b.t.w. why eliminate $\frac\pi 2$?), but (make a drawing with the trigonometric circle!): $$\theta+2k\pi<x <\pi-\theta+2k\pi\qquad(k\in\mathbf Z).$$

Similarly $$\sin x<\sin\theta \iff -\theta-\pi+2k\pi<x<\theta+2k\pi.$$