Question

Find general solution of the equations: cosec θ = - 2, cot θ = -√3

Answer 1

Case I :-

Given that

cosec θ = -2

The Reference Angle is π/6

cosec is -ve in Quadrant III & IV

For Quad III :-    θ  = π + π/6 = 7π/6

For Quad IV :-  θ = 2π - π/6 = 11π/6

∵    Period of cosec is  2π.

General Solution = { 7π/6 + 2nπ} U {11π/6 + 2nπ}

Case II :-

Given that

cot θ = [-\sqrt{3}[/tex]

The Reference Angle is π/6

cot is -ve in Quadrant II & IV

For Quad II :-    θ = π - π/6 = 5π/6

For Quad IV :-  θ = 2π - π/6 = 11π/6

∵    Period of cot is  π.

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

Answer 2

Solution:

To find the general solution of the equations: cosec θ = - 2, cot θ = -√3

$1) \: cosec \: \theta = - 2 \\ \\ \theta = {cosec}^{ - 1} ( - 2) \\ \\ \theta = {cosec}^{ - 1} (cosec( - \frac{\pi}{6} ))$
since principal value branch of
${cosec}^{ - 1} \: is \: [\frac{ - \pi}{2} ,\frac{\pi}{2} ] - (0)$
so,principal solution of the equation
$\theta = \frac{ - \pi}{6} \\ \\ x = ±\frac{\pi}{6} + 2\pi \: k$
where k is any integer

$2) \: let \: cot \: \theta = - \sqrt{3} = - cot \frac{\pi}{6} \\ \\ = cot(\pi - \frac{\pi}{6} ) \\ \\ = cot \: ( \frac{5\pi}{6} ) \\ \\ \theta = \frac{5\pi}{6}$

so,principal solution of the equation
$\theta = ± \frac{5\pi}{6} + \pi \: k$
where k is any integer