Discussion Topic

# Solving trigonometric equations involving tangent

Summary:

To solve trigonometric equations involving tangent, isolate the tangent function and then use the arctangent function to find the principal solution. Next, determine the general solution by adding integer multiples of the period, π, to the principal solution. This accounts for all possible solutions within the periodic nature of the tangent function.

Solve equationtan^2x-8tanx+12=0

We have to solve the equation: (tan x)^2 - 8 tan x + 12 = 0

let tan x = y

(tan x)^2 - 8 tan x + 12 = 0

=> y^2 - 8y + 12 = 0

=> y^2 - 6y - 2y + 12 = 0

=> y(y -6) - 2( y - 6) = 0

=> ( y - 2)(y - 6) = 0

So  y = 2 and y = 6

As y = tan x

tan x = 2 and tan x = 6

=> x = arc tan 2 + n*pi and x = arc tan 6 + n*pi

Therefore the solution is

x = arc tan 2 + n*pi and

x = arc tan 6 + n*pi

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Solve the equation tan^2x=8-8secx.

We have to solve (tan x)^2 = 8 - 8*sec x.

(tan x)^2 = 8 - 8*sec x

=> (sin x)^2 / (cos x)^2 = 8 - 8/(cos x)

=> (1 - (cos x)^2) / (cos x)^2 = (8*(cos x)^2 - 8* cos x)/ (cos x)^2

=> (1 - (cos x)^2) = (8*(cos x)^2 - 8* cos x)

=> 9(cos x)^2 - 8 cos x - 1 = 0

=> 9(cos x)^2 - 9 cos x  + cos x - 1 = 0

=> 9(cos x)(cos x - 1) + 1( cos x - 1) = 0

=> (9(cos x) - 1)(cos x - 1) = 0

cos x = 1 and cos x = 1/9

x = arc cos 1 and x = arc cos (1/9) and - arc cos (1/9)

The required values are : x = 2*n*pi and x = arc cos (1/9) + 2*n*pi and x = -arc cos (1/9) + 2*n*pi