Question

Solve 8sin2x – 6 = 0

Answer 1

Oops its suppose to be \[8\sin^2x-6=0\]

Answer 2

\(\large\color{black}{ \bf 8sin2x – 6 = 0 }\)
\(\large\color{black}{ \bf 8sin2x = 6 }\)
\(\large\color{black}{ \bf sin2x = 6/8 }\)
\(\large\color{black}{ \bf sin2x = 3/4 }\)
\(\large\color{black}{ \bf sin(x+x) = 3/4 }\)
\(\large\color{black}{ \bf sin(x)cos(x)+sin(x)cos(x) = 3/4 }\)
\(\large\color{black}{ \bf 2sin(x)cos(x) = 3/4 }\)
\(\large\color{black}{ \bf sin(x)cos(x) = 3/8 }\)
\(\large\color{black}{ \bf sin^2(x)cos^2(x) = 9/64 }\)
\(\large\color{black}{ \bf sin^2(x)(1-sin^2x) = 9/64 }\)
\(\large\color{black}{ \bf sin^2(x)-sin^4x = 9/64 }\)

Answer 3

Let sin^2x= a and solve for x.

Answer 4

I am so confused!

Answer 5

Which part do you get stuck at?

Answer 6

After doing solomon's work you need to solve for x from the trinomial