Where to go

To solve the angles for this equation where do you go?

sin2x + cosx = 0

all i can think of is

2sinxcosx + cosx =0

Where can you go from this?

Thanks

Reply 1

Gauss

Alex H

To solve the angles for this equation where do you go?

sin2x + cosx = 0

all i can think of is

2sinxcosx + cosx =0

Where can you go from this?

Thanks

yeh take the cosx out the bracket:

cosx(2sinx + 1) = 0 and solve for x

-> cosx = 0 => x= pi/2, 3pi/2, ...
-> 2sinx + 1 = 0 => sinx = -1/2 => x= -pi/6, 7pi/6, ...

Reply 2

Newton

2sixcosx+cosx=0

cosx(2sinx+1)=0

Now solve each part so that:

cosx=0 => x=(π/2)=90deg, (3π/2)=270deg

sinx=(-1/2) => x=(7π/6)=210deg, (11π/6)=330deg

Newton.

Reply 3

Alex H

Thanks, also

sin2x - 1 = cos2x

Any help appreciated

Thanks

Reply 4

BloodyValentine

sin2x=2sinxcosx as we've said
cos2x= 2sin^2 x -1 (sine squared)

bring it together

2sinxcosx - 2sin^2 x = 0
sinx(cosx-sinx)=0
sinx=0
Or sinx = cosx
which you can do

Reply 5

El Stevo

sin2x - 1 = cos2x
2sinxcosx - 1 = (cosx)^2 - (sinx)^2
2sinxcosx - 1 = (cosx)^2 - (1-(cosx)^2)
2sinxcosx - 1 = 2(cosx)^2 - 1
2(cosx)^2 + 2sinxcosx - 2 = 0
(cosx)^2 + sinxcosx - 1 = 0 ((cosx)^2 - 1 + sinxcosx = 0)
-(sinx)^2 + sinxcosx = 0
(sinx)^2 - sinxcosx = 0
sinx(sinx - cosx) = 0

sinx = 0, therefore x = 0, 180, 360

sinx - cosx = 0
sinx = cosx
tanx = 1, therefore, x = 45, 225

x = 0, 45, 180, 225, 360

someone better check my working though... my identities are pulled from somewhere in my head i thought i'd forgotten, and i tried to 'remember' the values... (no calc you see...)

edit: tpyoed wrong sign in second line... fixeded :smile: