Question

I will give brainliest if you answer properly.

Answer 1

Answer:

See below

Step-by-step explanation:

a)

$2\sin(x) +\sqrt{3} =0 \implies 2\sin(x)=-\sqrt{3} \implies \boxed{\sin(x)=-\dfrac{\sqrt{3}}{2} }$

$\therefore x=\dfrac{4\pi }{3}$

But note, as sine does represent the $y$ value, $\dfrac{5\pi }{3}$ is also solution

Therefore,

$x=\dfrac{4\pi }{3} \text{ and } x=\dfrac{5\pi }{3}$

This is the solution for $x\in[0, 2\pi ]$, recall the unit circle.

Note: $\sin(x)=-\dfrac{\sqrt{3}}{2} \implies \sin(x)=\sin \left(\pi +\dfrac{\pi }{3} \right)$

b)

$\sqrt{3} \tan(x) + 1 =0 \implies \tan(x) = -\dfrac{1}{\sqrt{3} } \implies \boxed{ \tan(x) = -\dfrac{\sqrt{3} }{3} }$

Once

$\tan(x) = -\dfrac{\sqrt{3} }{3} \implies \sin(x) = -\dfrac{1}{2} \text{ and } \cos(x) = \dfrac{\sqrt{3} }{2}$

As $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$

$\therefore x=-\dfrac{\pi }{6}$

c)

$4\sin^2(x) - 1 = 0 \implies \sin^2(x) = \dfrac{1}{4} \implies \boxed{\sin(x) = \pm \dfrac{\sqrt{1} }{\sqrt{4} } = \pm \dfrac{1}{2}}$

Therefore,

$\sin(x)=\dfrac{1}{2} \implies x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6}$

$\sin(x)=-\dfrac{1}{2} \implies x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}$

The solutions are

$x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6} \text{ and }x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}$