2cos^2(x) - cos(x) = 0?

anonymous

Solve for x if anyone knows how to do this i would love to know how thanks.

puggy

Your first step is to factor.

cos(x) [2cos(x) - 1] = 0

Now, equate each factor to 0

cos(x) = 0

2cos(x) - 1 = 0

cos(x) = 0

2cos(x) = 1

cos(x) = 0

cos(x) = 1/2

And then you solve. For the first equation,

x = pi/2, 3pi/2

For the second equation,

x = pi/3, 5pi/3

So your solutions, presuming that 0 <= x < 2pi, are

x = pi/2, 3pi/2, pi/3, 5pi/3

grant-phillips_609cad56de8f0

2cos 2x = 0 cos 2x = 0 2x = pi/2 + n* pi for the classic answer (that's, extraordinary values of npi/2, the position n is an integer) dividing by utilizing 2: x = pi/4 + npi/2 or, x = pi/4 (a million + 2n) <== established answer in case you only like the innovations between 0 and 2pi, then you definately'll may favor to allow n = 0 , a million , 2 , and three for 2 complete cycles for cos 2x in era [0 , 2pi), X = pi/4 , 3pi/4 , 5pi/4, and 7pi/4

anonymous

Just solve for cos(x):

2cos^2(x) = cos(x)

cos(x) = 0 and 1/2

x= 60 degrees +n*180 degrees

= 90 " +" "

anonymous

You can take out a common factor of cosx to give you

cosx(2cosx-1) = 0

cosx = 0 and 2cosx - 1 = 0

cosx = 0 and cos x = 1/2

Hence x = 60,90,270,300 all answers in degrees

maussy

put cos x =y

you find 2y^2 -y =0

y (2y-1) =0

two answers y= 0 y =+0.5

translating in angles y = (2k+1) 90° k = integer

or y = 60° + 2kpi

jim-burnell

u = cos x

2u² - u = 0

u (2u - 1) = 0

u = 0 or 2u - 1 = 0 so u = 1/2

cos x = 0

x = odd multiples of pi/2

cos x = 1/2

x = pi/3 (+/- 2npi) or -pi/e (+/- 2npi)

san

2cos^2(x) - cos(x) = 0

let u=cos(x)

then,

2u^2-u=0

u(2u-1)=0

therefore,

u=o ;or; u=1/2

when u=0 or when u=1/2

cos(x)=0 or cos(x)=1/2

(x)=90degree or (x) = 60degree

that all......cheers...

dupinder-jeet-kaur-k

cosx(2cosx-1)=0

cosx=0 or 2cosx-1=0

x=(2m+1)pi/2

or

cosx=1/2

x=2npi+pi/3,2npi-pi/3

inov8ed

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