Question

Solve the equation on the interval 0≤theta<2π.
(4cos^2)x - 3 ( I'll rewrite this so it makes more sense)

Answer 1

Solve the equation on the interval \[0\le \theta <2\pi \]Here's the equation:\[4\cos ^{2}x-3=0\]

Answer 2

try to isolate cos^2 x from there first, can you ?

Answer 3

cos^2x=3/4?

Answer 4

And welcome back (:

Answer 5

hey, thanks :)
ok, so cos^2x=3/4
take square root on both sides,
cos x= ... , ..... ?

Answer 6

(√3/4)? and yw (:

Answer 7

did you mean \(\large \sqrt{\dfrac{3}{4}}\)
?

Answer 8

Yup o:

Answer 9

also when you take square root , you get 2 values of cos x , you know what the other value is ?

Answer 10

hint : \(if x^2=a, \implies x=\pm \sqrt{a}\)

Answer 11

I think you lost me o;

Answer 12

\(\cos^2 x =3/4 \\ \implies \cos x = \pm \sqrt{3/4}= \sqrt 3/2, -\sqrt 3/2\)
got this ?

Answer 13

How'd you get ((√3)/2) from √3/4)?

Answer 14

\(\sqrt{3/4}= \sqrt3/\sqrt4\)
abd square root of 4 is 2
=\(\sqrt 3/2\)

Answer 15

*and

Answer 16

Oh, okay! Go on (:

Answer 17

for what values of x in unit circle 0≤theta<2π. is cos x = sqrt 3/2 ??
for what values of x in unit circle 0≤theta<2π. is cos x = -sqrt 3/2 ??

Answer 18

π/6, 5π/6, 7π/6, 11π/6 (:

Answer 19

Thank you!

Answer 20

\(\huge \checkmark\)
welcome ^_^