Question

Solve sin(4x)=-2sin(x)

...sin(2x+2x)+2sin(x)=0

Answer 1

Look up some double angle formulas..
What's Sin(2a)?

Answer 2

Sin(4x) = Sin(2*2x) = 2Sin(2x)Cos(2x) = -2Sin(x)

Answer 3

\[2(2Sin(x)*Cos(x))*(Cos^2(x) - Sin^2(x))=-2Sin(x)\]

Answer 4

Dividing both sides by \[2Sin(x)\]we get:\[2Cos(x)(Cos^2(x)-Sin^2(x))=-1\]

Answer 5

Aandd now I'm stuck.. im sure this is just some algebra and more simplification... hang on..

Answer 6

Well actually -1 = -(Sin^2 + Cos^2)

Answer 7

uhm yea that doesn't really help us anyway..

Answer 8

Looking for the solutions, which you should be able to get from the factors. Th answers should be 0, (pi/2), (pi), (3pi/2)

Answer 9

Wolfram is doing some crazy substitutions... i dont think they're necessary..

Answer 10

@hartnn @Zarkon

Answer 11

sin 4x = -2sin x
2 sin 2x cos 2x = -2 sin x
4 sin x cos x (2 cos^2x -1) + 2 sinx =0
factoring out 2 sin x
2 sin x [2 cos x (2cos^2x -1)+ 2 ] = 0
which gives us
sin x =0 OR 2 cos^3 x - cos x + 1 = 0

if we let cos x = p
we have a cubic function in p
2p^3 -p +1 = 0
solve this to get p = cos x = .... , ... , ...