Answer:
See below
Step-by-step explanation:
a)
[tex]2\sin(x) +\sqrt{3} =0 \implies 2\sin(x)=-\sqrt{3} \implies \boxed{\sin(x)=-\dfrac{\sqrt{3}}{2} }[/tex]
[tex]\therefore x=\dfrac{4\pi }{3}[/tex]
But note, as sine does represent the [tex]y[/tex] value, [tex]\dfrac{5\pi }{3}[/tex] is also solution
Therefore,
[tex]x=\dfrac{4\pi }{3} \text{ and } x=\dfrac{5\pi }{3}[/tex]
This is the solution for [tex]x\in[0, 2\pi ][/tex], recall the unit circle.
Note: [tex]\sin(x)=-\dfrac{\sqrt{3}}{2} \implies \sin(x)=\sin \left(\pi +\dfrac{\pi }{3} \right)[/tex]
b)
[tex]\sqrt{3} \tan(x) + 1 =0 \implies \tan(x) = -\dfrac{1}{\sqrt{3} } \implies \boxed{ \tan(x) = -\dfrac{\sqrt{3} }{3} }[/tex]
Once
[tex]\tan(x) = -\dfrac{\sqrt{3} }{3} \implies \sin(x) = -\dfrac{1}{2} \text{ and } \cos(x) = \dfrac{\sqrt{3} }{2}[/tex]
As [tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]
[tex]\therefore x=-\dfrac{\pi }{6}[/tex]
c)
[tex]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}}[/tex]
Therefore,
[tex]\sin(x)=\dfrac{1}{2} \implies x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6}[/tex]
[tex]\sin(x)=-\dfrac{1}{2} \implies x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]
The solutions are
[tex]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}[/tex]
