Question

Solve for variable involving trig functions

Answer 1

Adding equation with work.

Answer 2

\[m \cos x+m \sin x=\cos 2x\]

Answer 3

Solve for m.

Answer 4

\[m \cos x+msinx=\cos ^2x-\sin ^2x\]
\[\frac{ m \cos x+m \sin x }{ cosx+sinx }=\frac{ \cos^2x-\sin^2x }{cosx+sinx }\]
\[m^2=cosx-sinx\]
\[m=\sqrt{cosx-sinx}\]

Answer 5

I'm pretty sure that's not right...

Answer 6

why is \(m^2\) equal to cos x- sin x
wouldn't it be just m = cos x -sin x

Answer 7

ab+ac = a(b+c)
m cos x- m sin x = m (cos x-sin x)

Answer 8

That makes sense! Thanks.

Answer 9

welcome ^_^