Find the general solution (i) $\cot \theta =-\dfrac {1} {\sqrt {3}}$ (ii)$4\cos ^{2}\theta =1$

Question

my attempt for (i)

$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$

$\cot ( \theta ) = - \frac { \sqrt { 3 } } { 3 }$

(ii)

$\left. \begin{array} { l } { \text { Let: } \cos ( \theta ) = u } \\ { 4 u ^ { 2 } = 1 } \end{array} \right.$

$\left. \begin{array} { l } { \text { Divide both sides by } 4 } \\ { \frac { 4 u ^ { 2 } } { 4 } = \frac { 1 } { 4 } } \end{array} \right.$

is it right way to find general solution for these equations?

Answer 1

We know that $$\cot\left(-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt 3}{3}$$ and $$\cot( \theta+\pi)=\cot \theta$$similarly $$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$and $$\cos^2(\theta+\pi)=\cos^2\theta$$therefore $$(i)\qquad x=k\pi-\dfrac{\pi}{3}\\(ii)\qquad x=k\pi\pm\dfrac{\pi}{3}$$

Answer 2

$$1=4\cos^2t=2(1+\cos2t)$$

$$\iff\cos2t=?$$

$$2t=2n\pi\pm\dfrac{2\pi}3$$