Question

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

Answer 1

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)`.