Question

2 sin 3theta -sqrt2 = 0

Answer 1

\[2\sin 3\theta-\sqrt{2}=0=>2\sin 3\theta=\sqrt{2}=>\sin 3\theta=\frac{ 1 }{ \sqrt{2} }=>3\theta=\sin^{-1} \left( \frac{ 1 }{ \sqrt{2} } \right)=>3\theta=45=>\theta=15\]

Answer 2

2 sin 3theta -sqrt2 = 0
2 sin(3theta)=sqrt(2)
sin(3 theta) = sqrt(2)/2
3theta = arcsin(sqrt(2)/2)
theta = arcsin(sqrt(2)/2)/3
theta = (PI/4 + 2PI*n)/3 where n belongs to Z
theta = PI/12 + 2PI*n/3 where n belongs to Z

Answer 3

\[3\theta=45=>\theta=15 (answer)\]

Answer 4

you typed it in degrees however sin has a period of 2PI thus there is an infinite set of answers

Answer 5

but here its not mentioned that in which form find the answer..!!!if there is any restricton then it colud sooo..1!!

Answer 6

Exactly. No bounds specified then you need to show all possible answers.

Answer 7

okay..!!