How do you solve #sin(2x)=sqrt3/2# and find all solutions in the interval #[0,2pi)#?

1 Answer
Ratnaker Mehta
Jul 26, 2016

The Soln. Set #={pi/6, pi/3, 7pi/6, 4pi/3} sub [0, 2pi)#.

Explanation:

#sin(2x)=sqrt3/2=sin(pi/3)..............(1)#

Now, let use the soln. for the trigo. eqn.

# : sintheta=sinalpha rArr theta=kpi+(-1)^kalpha, k in ZZ#

Hence, using this soln. for #(1)#, we get,

#2x=kpi+(-1)^k(pi/3), or, x=kpi/2+pi/6(-1)^ k, in ZZ#

#k=-2, x=-pi+pi/6=-5pi/6 !in [o,2pi)=I, say.#
#k=-1, x=-pi/2-pi/6=-2pi/3 !in I#
#k=0, x=pi/6 in I#
#k=1, x=pi/2-pi/6=pi/3 in I#
#k=2, x=pi+pi/6=7pi/6 in I#
#k=3, x=3pi/2-pi/6=4pi/3 in I#
#k=4, x=2pi+pi/6=13pi/6 !in I#

#:.# The Soln. Set #={pi/6, pi/3, 7pi/6, 4pi/3} sub [0, 2pi)#.

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