##### Solve each equation over the interval [ 0 , 2pi).

 Calculus Tutor: None Selected Time limit: 1 Day

Solve each equation over the interval [ 0 , 2pi).

1. 2 Tan x Sin x – Sec x = 2 Cos x

May 21st, 2015

There are multiple ways to do this. You just have to play around with it. Here's one possibility:
Expand tan (2x) using the double angle formula
tan (2x) - tan(x) = 0
(2 tan x) / (1 - tan^2 x) - tan x = 0
factor out tan x
tan x ( 2/(1- tan^2 x) -1 ) = 0
convert tan^2 x to sin^2 x / cos^2 x, then multiply top and bottom of fraction by cos^2 x
tan x [ (cos^2 x/(cos^2 x - sin^2 x) -1] = 0
substitute for 1 = (cos^2 x - sin^2 x)/(cos^2 x - sin^2 x) and combine the 2 terms
tan x [ {2 cos^2 x - cos^2 x + sin^2 x)/ ((cos^2 x - sin^2 x)
Simplify the numerator
tan x [ cos^2 x + sin^2 x] / (cos^2 x - sin^2 x)
tan x / (cos^2 x - sin^2 x) = 0
Use one more double angle formula, cos (2x) = cos^2 x - sin^2 x
tan x / cos (2x) = 0
Or,
tan x * sec (2x) = 0
This is true only when tan x = 0 or sec (2x) = 0, but sec x is never 0. So
tan x = 0 when x = pi * n, for integer n.

May 21st, 2015

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May 21st, 2015
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May 21st, 2015
Dec 8th, 2016
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