Answer:
[tex]\sf\:Negative = \boxed{\sf\:\theta =\pi n_{1}+\frac{5\pi }{6}\text{, }n_{1}\in \mathbb{Z}}\\\sf\: Positive = \boxed{\sf\:\theta =\pi n_{1}+\frac{\pi }{6}\text{, }n_{1}\in \mathbb{Z}}[/tex]
Step-by-step explanation:
[tex]4 - 3 \sec ^ { 2 } \theta = 0[/tex]
Let's solve this.
Isolate sec²θ.
[tex]4 -3 \sec ^ { 2 } \theta = 0\\- 3 \sec^{2}\theta = -4\\3 \sec^{2}\theta=0\\\sec^{2} \theta=\frac{4}{3}[/tex]
Now, we now that, cos θ is the reciprocal of sec θ. Therefore,
[tex]\sec^{2}\theta = \frac{4}{3}\\\cos^{2}\theta = \frac{3}{4}[/tex]
Bring the square root on both the sides of the equation to remove the square.
[tex]\sqrt{\cos^{2}\theta} = \sqrt{\frac{3}{4} }\\\cos^{2}\theta = (+/-) \frac{\sqrt{3} }{2}[/tex]
If,
[tex]\cos\theta = + \frac{\sqrt{3}}{2}\\= cos \left( \frac{\pi}{6} \right)\\\Longrightarrow \boxed{\sf\:\theta =\pi n_{1}+\frac{\pi }{6}\text{, }n_{1}\in \mathbb{Z}}[/tex]
Also if,
[tex]\cos\theta = -\frac{\sqrt{3}}{2}\\= \cos \left(\pi - \frac{\pi}{6}\right)\\= \cos \left({\frac{5\:\pi}{6}\right)\\\Longrightarrow \boxed{\sf\:\theta =\pi n_{1}+\frac{5\pi }{6}\text{, }n_{1}\in \mathbb{Z}}[/tex]
[tex]\rule{150pt}{2pt}[/tex]
