Question #21463
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"Question #a738e"
For #0<=theta<2pi#, #theta = pi/5, (5pi)/6, (3pi)/2#
#cos^2theta-sin^2theta=sintheta#
Use the trig identity #sin^2theta + cos^2theta =1#
#=> cos^2theta=1-sin^2theta#
Substitute for #cos^2theta#
#1-sin^2theta-sin^2theta=sintheta#
#-2sin^2theta +1=sintheta#
#0=2sin^2theta+sintheta-1#
Factor
#(2sintheta-1)(sintheta+1)=0#
Set each factor equal to zero and solve.
#2sintheta - 1 =0 color(white)(aaa)sintheta+1=0#
#sintheta=1/2 color(white)(aaa)sintheta=-1#
For #0<=theta<2pi#, use unit circle to find
#theta =pi/6, (5pi)/6color(white)(aaa)theta =(3pi)/2#
#cos^2theta-sin^2theta=sintheta#
#=>cos2theta=cos(pi/2-theta)#
#2theta=2npi±(pi/2-theta)," where "n in ZZ#
When #2theta=2npi+(pi/2-theta)#
#=>3theta=(4n+1)pi/2#
#=>theta=(4n+1)pi/6#
Again when
#2theta=2npi-(pi/2-theta)#
#=>2theta=2npi-pi/2+theta#
#=>theta=(4n-1)pi/2#
