How do you solve tan^2x/secx+cosx=2 for 0<=x<=2pi?

1 Answer
Apr 19, 2018

x=pi/3, (5pi)/3

Explanation:

First, recalling that tan^2=sin^2x/cos^2x, secx=1/cosx, simplify the equation:

(sin^2x/cos^2x)/(1/cosx)+cosx=2

sin^2x/(cos^cancel(2)x/cancelcosx)

sin^2x/cosx+cosx=2

Add the expressions, using cosx as a common denominator:

(sin^2+cos^2x)/cosx=2

Recall that sin^2x+cos^2x=1:

1/cosx=2

Invert both sides:

cosx=1/2

In the interval [0, 2pi], we examine the unit circle and see this holds true for x=pi/3, (5pi)/3

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