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
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
Thats not correct....
I'm not sure how you got from
2 - 2cos^2(x) - 5cos(x) + 1 = 0
2cos^2(x) + 5cos(x) + 3 = 0
It should be -2cos^2(x) - 5cos(x)+1=0
From there I'm stuck.
sorry. meant
-2cos^2(x) - 5cos(x)+3=0
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
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)
Thank you....the pi/3, 5pi/3 is correct.
To solve the given equation in the interval [0, 2π], we can use the following steps:
Step 1: Rewrite the equation in terms of sin(x).
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
Step 2: Let u = cos(x). Now the equation becomes a quadratic equation in terms of u.
-2u^2 - 5u + 3 = 0
Step 3: Solve the quadratic equation.
To solve the quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:
u = (-b ± √(b^2 - 4ac))/(2a)
Here, a = -2, b = -5, and c = 3.
u = (-(-5) ± √((-5)^2 - 4(-2)(3))) / (2(-2))
u = (5 ± √(25 + 24)) / (-4)
u = (5 ± √(49)) / (-4)
u = (5 ± 7) / (-4)
u = (5 + 7) / (-4) or u = (5 - 7) / (-4)
u = 12 / -4 or u = -2 / -4
u = -3 or u = 1/2
Step 4: Find x using the values of u.
Since u = cos(x), we can now solve for x.
When u = -3:
cos(x) = -3
However, the range of the cosine function is [-1, 1], which means there is no solution for cos(x) = -3 in the interval [0, 2π].
When u = 1/2:
cos(x) = 1/2
To find the value(s) of x, we can use the inverse cosine function (arccos). Since the answer must be written as a multiple of π, we have:
x = π/3 or x = 5π/3
Therefore, the solutions to the equation 2(sin(x))^2 - 5cos(x) + 1 = 0 in the interval [0, 2π] are:
x = π/3, 5π/3 (written as multiples of π).