How do you find the exact value of #cos2theta+2cos^2theta=2# in the interval #0<=theta<360#?

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Answer 1 · 2016-12-16T16:18:04

The solutions are #theta={30º,150º,210º,330º}#

We use #cos2theta=2cos^2theta-1#

Therefore our equation is

#cos2theta+2cos^2theta-2=0#

#2cos^2theta-1+2cos^2theta-2=0#

#4cos^2theta=3#

#cos^2theta=3/4#

#cos theta=+-sqrt3/2#

First, #costheta=sqrt3/2#, #=>#, #theta=30º ; 330º#

Second, #costheta=-sqrt3/2#, #=>#, #theta=150º; 210º#

Answer 2 · 2016-12-16T16:27:58

Please see the explanation

Given: #cos(2theta) + 2cos^2(theta) = 2; 0^@ le theta < 360^@#

The identity #cos(2theta) = 2cos^2(theta) - 1# allows allows us to substitute #2cos^2(theta) - 1# for #cos(2theta)#:

#2cos^2(theta) - 1 + 2cos^2(theta) = 2#

Combine like terms:

#4cos^2(theta) = 3#

Divide by 4:

#cos^2(theta) = 3/4#

square root both sides:

#cos(theta) = +-sqrt(3)/2#

The angles are well known:

#theta = 30^@, 150^@, 210^@, and 330^@#