Solving Trigonomic Equation. IF 0<x<360, what is the value of cosx in the equation 2cos^2x+sin^2x=41/25?

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Answer 1 · 2018-01-06T18:20:53

#cos(x) = pm 4/5#

We know that #cos^2(x)+sin^2(x)=1\rightarrow sin^2(x) = 1-cos^2(x)#.

Substituting we have:

#2cos^2(x)+(1-cos^2(x))=41/25#

#cos^2(x)+1=41/25#

#cos^2(x)=16/25#

#cos(x) = pm 4/5#

Since #0< x < 360^circ#, we can't rule out either value of #cos(x)#.