Question

Check my answer please? (Trig)

Answer 1

x= 2pi/3, pi/2, 3pi/2, and pi/3

Answer 2

pi/3 doesn't work

Answer 3

Ok are there any other solutions to it? and can you help me with the general formulae?

Answer 4

\[\sin^2x+\cos^2x=1\]so
\[\sin^2x=1-\cos^2x\]

Answer 5

\[2(1-\cos^2 x)=2+\cos x\\
2-2\cos^2 x=2+\cos x\\
0=\cos x +2\cos^2 x\\
0=\cos x(1+2\cos x)\]

so either

\[\cos x=0\] or\[1+2\cos x=0\]

Answer 6

i mean the general form for the values of x

Answer 7

like pi/2 + n pi

Answer 8

that condition solves this bit cos x=0

Answer 9

but what about 1+2cos x=0 ?

Answer 10

pi/2+2n pi?

Answer 11

1+2cos x=0
2cos x=-1
cos x=-1/2

Answer 12

I'm confused :S

Answer 13

i have all that, i'm looking for the general formula of my answers

Answer 14

well your almost there

Answer 15

for the first bit
cos x=0

you found
x= pi/2 + n pi

because
x= acos 0 = pi/2 + n pi

Answer 16

cos x=-1/2

x=acos(-1/2)

Answer 17

x=120?

Answer 18

yeahs, that solves it, but its not the general form

Answer 19

Ok so what is?

Answer 20

for which angles x, is cos x =-1/2