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

Question

3793 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-19T03:25:52

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

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#

enter image source here