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How do you solve #cosx+sinxtanx=2#?

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May 25, 2015

#tan(x) = sin(x)/cos(x)#

So
#cos(x) + sin(x)tan(x) = 2#

#rarr cos(x) + sin^2(x)/cos(x) = 2#

#rarr (cos^2(x) + sin^2(x))/cos(x) = 2#

#rarr 1/cos(x) = 2#

#rarr cos(x) = 1/2#

Based on one of the standard trigonometric triangles
#x = +-60^@#

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