solve 2cos^2x-5cosx+2=0 for principal values of x
I don't understand this question...what are principal values?
those solutions where 0 ≤ x ≤ 2π
2cos^2x-5cosx+2=0
this is a standard quadratic where your variable is cosx
How about doing this ?
let t = cosx
then we have
2t^2 - 5t + 2 = 0 , which factors to
(2t - 1)(t - 2) = 0
t = 1/2 or t = 2
so cosx = 1/2 or cosx = 2
that last part is silly, since cos(anything) lies between -1 and +1
so cosx = +1/2, which tells me that the angle x must be in quads I or IV by the CAST rule
we know that x = 60° or π/3
We are in I or IV
so x = 60 = 60° or π/3
or x = 360-60 = 300° or 5π/3
In trigonometry, principal values refer to the values of an angle that lie within a specific range. For example, in the case of the cosine function, the principal values typically lie between 0 and 2π radians (or 0 and 360 degrees).
To solve the equation 2cos^2x - 5cosx + 2 = 0 for the principal values of x, we can use the quadratic formula. However, since the equation involves cosine, we need to rewrite it in terms of a quadratic equation. Let's start the process:
The equation is: 2cos^2x - 5cosx + 2 = 0
We can rewrite cos^2x as (1 - sin^2x) using the identity cos^2x = 1 - sin^2x:
2(1 - sin^2x) - 5cosx + 2 = 0
2 - 2sin^2x - 5cosx + 2 = 0
Rearrange the terms:
-2sin^2x - 5cosx + 4 = 0
Now we can solve this quadratic equation using the quadratic formula:
sinx = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = -2, b = -5, and c = 4:
sinx = (5 ± √((-5)^2 - 4(-2)(4))) / (2(-2))
sinx = (5 ± √(25 + 32)) / (-4)
sinx = (5 ± √57) / (-4)
Now, to find the corresponding values of x, we can use the inverse sine function (sin^(-1)) and plug in the values of (5 ± √57) / (-4):
x = sin^(-1)((5 + √57) / (-4))
x = sin^(-1)((5 - √57) / (-4))
Using a calculator or table, we can evaluate the inverse sine. Keep in mind that we are looking for the principal values of x, so we will only consider the angles that fall within the range of 0 to 2π radians or 0 to 360 degrees.
In mathematics, the principal values of an equation typically refer to the solutions that lie within a specific range or interval. For trigonometric functions such as cosine, the principal values of an angle usually lie between 0 and 2π (or 0 and 360 degrees).
Now, let's solve the equation 2cos^2x - 5cosx + 2 = 0 for the principal values of x.
To do this, we can use a substitution. Let's set cosx as a variable, such as y. So, we have the equation 2y^2 - 5y + 2 = 0.
Now, we can factorize this quadratic equation. We need to find two numbers that multiply to give 2ac (where a = 2, b = -5, and c = 2) and add up to give -b.
In this case, 2 * 2 = 4, and we need to find two numbers that multiply to give 4 and add up to give -5. These numbers are -4 and -1.
So, we can rewrite the equation as (2y - 1)(y - 2) = 0.
Now, we solve each factor separately:
Setting 2y - 1 = 0, we find y = 1/2. Since cosx = y, we get cosx = 1/2.
To find the angle x, we can take the inverse cosine (cos⁻¹) of 1/2, which gives x = π/3 (or 60 degrees since π radians is equivalent to 180 degrees).
Setting y - 2 = 0, we find y = 2. Since cosx = y, we get cosx = 2.
However, the value of cosine cannot be greater than 1 or less than -1, so there are no solutions for this case.
Therefore, the principal value of x for the given equation 2cos^2x - 5cosx + 2 = 0 is x = π/3 (or 60 degrees).