How do you solve #\sin 3\theta - \sin \theta = 4\cos ^ { 2} \theta - 2#?

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Answer 1 · 2017-08-28T08:52:40

#:."The Soln. Set="{kpi+(-1)^k(pi/2)}uu{kpi+(-1)^k(+-pi/4)}, k in ZZ.#

Knowing that, #sin3theta=3sintheta-4sin^3theta,# we have,

#(3sintheta-4sin^3theta)-sintheta=4(1-sin^2theta)-2.#

#:. 2sintheta-4sin^3theta=4-2-4sin^2theta, i.e., #

# 2sintheta(1-2sin^2theta)-2(1-2sin^2theta)=0,#

#:. 2(sintheta-1)(1-2sin^2theta)=0.#

#:. sintheta=1=sin(pi/2), or, sintheta=pm1/sqrt2=sin(pmpi/4).#

Since, #sinx=siny rArr x=kpi+(-1)^ky, k in ZZ,# we have,

#theta=kpi+(-1)^k(pi/2), or, kpi+(-1)^k(+-pi/4), k in ZZ.#

#:."The Soln. Set="{kpi+(-1)^k(pi/2)}uu{kpi+(-1)^k(+-pi/4)}, k in ZZ.#

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