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C3 solving trigs question watch

1. I this looks easy, but how do I solve:

sin2x = sinx (0< x < 360)

I don't know how to re-aarange it

also....
how do I re-arrange: 8sin^3 x - 6sinx + 1 = 0, to solve it.

use the double angle formula

sin2x=2sinxcosx
so you've got 2sinxcosx=sinx
Then divide both sides by sinx
2cosx=1
cosx=1/2
And then solve it between 0 and 360.
3. sin2x = 2sinxcosx

therefore, 2sinxcosx = sinx

2sinxcosx - sinx = 0

sinx(2cosx-1) = 0

sinx = 0,
2cosx=1, cosx=1/2

i think thats right...
4. Your first one is a double angle formula.
sinx(2x)=2sinxcosx then work on from there...

The second one, let sinx = X then you get the cubic
8X^3 -6X +1=0, find the roots of it, which are equal to sinx. arcsin the roots then you have the solutions.
5. (Original post by Manitude)

use the double angle formula

sin2x=2sinxcosx
so you've got 2sinxcosx=sinx
Then divide both sides by sinx
2cosx=1
cosx=1/2
And then solve it between 0 and 360.
You actually don't want to divide by sinx, you lose solutions like that.

Just factorise it sinx(2cosx-1)= 0 and solve for sinx =0 and cosx =1/2
6. (Original post by Manitude)

use the double angle formula

sin2x=2sinxcosx
so you've got 2sinxcosx=sinx
Then divide both sides by sinx
2cosx=1
cosx=1/2
And then solve it between 0 and 360.
You lose a solution by dividing by sinx..

you need to make it 2sinxcosx - sinx = 0 and then factorise that to get all solutions..
7. yeah...good point, Clarity Incognito and andrew.P...I should have noticed that.

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