Question

Find general solution of the equations: tan x = $-\frac{1}{\sqrt{3}}$, sec x = $\frac{2}{\sqrt{3}}$

Answer 1

Case I :-

Given that

tan x =   $-\frac{1}{\sqrt{3} }$

The Reference Angle is π/6

Tan is -ve in Quadrant II & IV

For Quad II :-    x = π - π/6 = 5π/6

For Quad IV :-  x = 2π - π/6 = 11π/6

∵    Period of tan is  π.

General Solution = { 5π/6 + nπ} U {11π/6 + nπ}

Case II :-

Given that

sec x = $\frac{2}{\sqrt{3} }$

The Reference Angle is π/6

Tan is +ve in Quadrant I & IV

For Quad I :-    x = π/6

For Quad IV :-  x = 2π - π/6 = 11π/6

∵    Period of sec is 2 π.

General Solution = { π + 2nπ} U {11π/6 + 2nπ}