Question

Given that
3sin^2-8cosx-7=0
show that, for real values of x
cos x = -2/3

I'm really bad at trig identities so if anyone could help it would be great appreciated :D

Answer 1

did u know that \(\sin^2x+\cos^2x=1\)
?

Answer 2

Yes I do

Answer 3

so, from that what does sin^2 x = ... ?

Answer 4

1-cos^2x

Answer 5

good!
so, plug in 1-cos^2 x in place of sin^2 x in
3sin^2 x-8cos x -7=0
what u get ?

Answer 6

3(1-cos^2)-8cosx-7=0?

Answer 7

simplify that a bit ?
to bring it in the form of \(ax^2+bx+c=0\)
?

Answer 8

3-3cos^2-8cosx-7=0
=> -3cos^2-8cos-4=0

Answer 9

cool, lets put cos x = X for simplicity
can you solve this quadratic ?
-3X^2 -8X-4 =0 ???

or
3X^2+8X+4=0 ?

Answer 10

(-3x-2)(x-2)=0?

Answer 11

absolutely correct! good work :)

Answer 12

so, x was cos x
-3cos x-2 = 0
gives you what ?

Answer 13

-3cosx=2
=> cosx=-2/3
Thanks so much you make the questions so much easier for me :)

Answer 14

cos x+2 =0
gives cos x = -2
but the range of cos is between 1 and -1
so for no value of x. cos x will be -2

Answer 15

and you are welcome ^_^

Answer 16

Thanks!