Posted by **Bryan** on Friday, November 30, 2007 at 6:38pm.

Solve the given equation in the interval [0,2 pi].

Note: The answer must be written as a multiple of pi. Give exact

answers. Do not use decimal numbers. The answer must

an integer or a fraction. Note that pi is already provided with the

answer so you just have to find the appropriate multiple. E.g.

the answer is p/2 you should enter 1/2. If there is more than one

answer write them separated by commas.

2(sin(x))^2−5cos(x)+1 = 0

x=__________ pi

- Math -
**Reiny**, Friday, November 30, 2007 at 7:33pm
2(sin(x))^2−5cos(x)+1 = 0

2(1- cos^2(x))−5cos(x)+1 = 0

2 - 2cos^2(x) - 5cos(x) + 1 = 0

2cos^2(x) + 5cos(x) + 3 = 0

(cosx + 1 )(2cosx + 3) = 0

cosx = -1 or cosx = -3/2

from the first part x = 0 or 2pi

there is no solution to the second part since the cosine function cannot exceed ±1

- Math -
**Bryan**, Friday, November 30, 2007 at 7:46pm
sorry. meant

-2cos^2(x) - 5cos(x)+3=0

- Math -
**Reiny**, Friday, November 30, 2007 at 7:56pm
Oh my, OH my, two errors in the same post, I must be getting careless.

Let's start from

2(1- cos^2(x))−5cos(x)+1 = 0

2 - 2cos^2(x) - 5cos(x) + 1 = 0

2cos^2(x) + 5cos(x) - 3 = 0

(cosx + 3)(2cosx - 1) = 0

cosx = -3 which will not produce an answer

or

cosx = 1/2

so x = pi/3 or 5pi/3 (60º or 300º)

(these answers are correct, I tested them)

- Math -
**Reiny**, Friday, November 30, 2007 at 7:49pm
Thanks for catching my error,

of course if cos x = -1 then x = pi.

I read my own writing as cosx = 1, (let's blame it on eyesight of an old guy, lol)

you said:"I'm not sure how you got from

2 - 2cos^2(x) - 5cos(x) + 1 = 0 "

look at 2(1- cos^2(x)) and expand it, you get

2 - 2cos^x

so x = pi is your only solution

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