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Find the general value of (theta) satisfying: 1) sin2(theta)=1/2 2) cos2(theta)=1/2 3) tan3(theta)=1/√3 4) cos3(theta)=-√3/2 5) sin²(theta)=3/4 6) sin²2(theta)=1/4 7) 4cos²(theta)=1 8) cos²2(theta)=3/4??

Answer 1

Elijah Alexander

Please see below.

(1) #sin2theta=1/2=sin(pi/6)# hence #2theta=npi+(-1)^npi/6#

or #theta=(npi)/2+(-1)^npi/12#

(2) #cos2theta=1/2=cos(pi/3)# hence #2theta=2npi+-pi/3#

or #theta=npi+-pi/6#

(3) #tan3theta=1/sqrt3=tan(pi/6) hence #3theta=npi+pi/6#

or #theta=(npi)/3+pi/18#

(4) #cos3theta=-sqrt3/2=cos((5pi)/6)# hence #3theta=2npi+-(5pi)/6#

or #theta=(2npi)/3+-(5pi)/18#

(5) #sin^2theta=3/4=sin^2(pi/3)# hence #theta=npi+-pi/3#

(6) #sin^2 2theta=1/4=sin^2(pi/6)# hence #2theta=npi+-pi/6#

or #theta=(npi)/2+-pi/12#

(7) #4cos²theta=1=>cos^2theta=1/4=cos^2(pi/3)# hence #2theta=npi+-pi/3# or #theta=(npi)/2+-pi/6#

(8) #cos^2 2theta=3/4=cos^2(pi/6)# hence #2theta=npi+-pi/6#

or #theta=(npi)/2+-pi/12#

Here #ninZZ#

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