How do you solve #2-2sintheta=4cos^2theta#?

Question

What is the solution in the interval #0<=theta<2pi#?

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Answer 1 · 2017-01-07T22:00:15

In the interval #(-pi,pi]# the equation is solved for:

#theta = -(5pi)/6#
#theta = -pi/6#
#theta = pi/2#

As #cos^theta = 1- sin^2theta# we have:

#2-2sintheta = 4cos^2 theta#

#1-sin theta = 2(1-sin^2theta)#

#1-sin theta = 2-2sin^2theta#

#2sin^2theta-sin theta -1 =0#

Solving as a second order equation:

#sin theta = frac (1+- sqrt(1+8)) 4 = frac (1+-3) 4#

or

#sin theta_1 = -1/2#

#sin theta_2 = 1#

So in the interval #(-pi,pi]# the equation is solved for:

#theta = -(5pi)/6#
#theta = -pi/6#
#theta = pi/2#