Question

please help
2cos(^2)2x=-1+3cos2x

Answer 1

let \(\large y=cos(2x) \)
so that trig equation can be written as:
\(\large 2y^2=-1+3y \) or
\(\large 2y^2-3y+1=0 \)

can you solve for y?

Answer 2

so i plug the value for y into one of the equations?

Answer 3

you realize that the trig equation you have is actually a trig equation in quadratic form???
so you would solve the quadratic equation by factoring:

\(\large 2y^2-3y+1=0 \)
\(\large (2y-1)(y-1)=0 \)

so y = ????

Answer 4

but its \[2\cos ^{2}2\theta=-1+3\cos 2\theta\]

Answer 5

n dik wat 2 do w/the 2 in front of the theta

Answer 6

read what i said...
the trigonometric equation you have is a trig equation in QUADRATIC form....
to solve that trig equation, you would use the technique used to solve the quadratic equation: \(\large 2y^2=-1+3y \)

Answer 7

so if i use the quadratic formula then i shud b able to solve for y?

Answer 8

ok... this is your equation:
\(\large 2\cdot \color {red}{cos^22\theta}=-1+3\cdot \color {red}{cos2\theta} \)
\(\large 2\cdot \color {red}{y^2}=-1+3\cdot \color {red}{y} \)

sure you can use the quadratic formula but it would be easier to factor as i did earlier...

Answer 9

is y \[(-3\pm sort{17})/4\]

Answer 10

no... rewrite the quadratic in standard form, then factor:
\(\large 2y^2=-1+3y \)
\(\large 2y^2-3y+1=0 \) standard form
\(\large (2y-1)(y-1)=0 \) factor
so
\(\large 2y-1=0 \) or \(\large y-1=0 \)

now it's obvious y=1/2 , y = 1....

Answer 11

n y stands for cos2theta?

Answer 12

now that you have y=1/2 or y=1,
remember that we replaced cos22θ=y
so now you have to solve these two equations:
\(\large cos2\theta=\frac{1}{2} \) ; \(\large cos2\theta=1 \)

Answer 13

yes... y=cos2theta...

Answer 14

so its 30

Answer 15

n 0

Answer 16

you have several answers between [0, 2pi)

Answer 17

so 30 150 210 n 330?

Answer 18

or just is it just 30 n 330 since we are dealing with cos?

Answer 19

i only see 30, 330, 0 degrees

Answer 20

well, the 30 and 330 degrees came from the first equation: \(\large cos2\theta=\frac{1}{2} \)

and the 0 degrees came from cos2theta = 1

Answer 21

thank u!!!!!!! u helped so much.....by any chance cud u help me solve this next one?
6tanθ−6cotθ=0

Answer 22

yw...
post it up as a new question...

Answer 23

okie dokie :)