Question

Find the principal and general solutions of the equation $\sec x=2$

Answer 1

Step $1:$ Solve for principal solution.

Given $\sec x=2\Rightarrow \frac{1}{\cos x}=2\Rightarrow \cos x=\frac{1}{2}$

We know that $\cos {60}^{\circ }=\frac{1}{2}$
We find value of $x$ where $\cos$ is positive.
$\cos$ is positive in $1^{st}$ and $4^{th}$ quadrant.

Value in $1^{st}$ quadrant $={60}^{\circ }$
Value in $4^{th}$ quadrant $={360}^{\circ }-{60}^{\circ }={300}^{\circ }$
So, principal solutions are $x={60}^{\circ }=60\times \frac{\pi }{180}=\frac{\pi }{3}$ and $x={300}^{\circ }=300\times \frac{\pi }{180}=\frac{5\pi }{3}$

Step 2: Solve for general solution
$\cos x=\frac{1}{2}$
$\Rightarrow \cos x=\cos \frac{\pi }{3}$

We know that if $\cos x=\cos y$
General solution is $x=2n\pi \pm y$ where $n\in Z$
Put $y=\frac{\pi }{3}$
Hence, $x=2n\pi \pm \frac{\pi }{3}$ where $n\in Z$