Question

Find all solutions of the equation 2sinx +√3 = 0

The answer is A+Bkπ and C+Dkπ where k is any integer, 0

Answer 1

Explanation:

2sin x=-√3

`sin~x=-(√(3) )/(2) =-sin~(\pi )/(3) =sin (\pi +(\pi )/(3) ),sin(2\pi -(\pi )/(3) )\\=sin((4\pi )/(3) +2k\pi ),sin ((5\pi )/(3) +2k\pi )\\=sin(A+Bk\pi ),sin(C+Dk\pi )\\x=A+Bk\pi ,C+Dk\pi \\where~A=(4\pi )/(3) ,B=2\\C=(5\pi )/(3) ,D=2`