solve this equation on the given interval expressing the solution for x in terms of inverse trigonometric functions.
6(cosx)^2-cosx-5=0 on (pie/2, pie)
Isn't this a quadratic in cosx ?
6y^2-y-5=0
y=(1+-sqrt(1+120))/12
where y= cos x
then solve for x
by formula
cosx = (2 ± √124)/12
= -.76129 or 1.09.. , the latter is undefined
x = 2.436 radians
or x = cos^-1 ( (2-√124)/12)
I misread your equation, should be
x = cos^-1 ( 1 ± √121)/12)
= cos^-1 (-5/6)
To solve the equation 6(cosx)^2 - cosx - 5 = 0 on the given interval (∏/2, ∏) and express the solution for x in terms of inverse trigonometric functions, follow these steps:
Step 1: Rewrite the equation in terms of the cosine function.
6cos^2(x) - cos(x) - 5 = 0
Step 2: Simplify the equation.
Let's denote cos(x) as a variable t:
6t^2 - t - 5 = 0
Step 3: Solve the quadratic equation.
To solve the quadratic equation, we can use the quadratic formula. The quadratic formula is given as:
t = (-b ± √(b^2 - 4ac)) / (2a)
Coefficients:
a = 6, b = -1, c = -5
Plugging in the values:
t = (-(-1) ± √((-1)^2 - 4(6)(-5))) / (2(6))
= (1 ± √(1 + 120)) / 12
= (1 ± √121) / 12
= (1 ± 11) / 12
Simplifying:
t1 = (1 + 11) / 12 = 12 / 12 = 1
t2 = (1 - 11) / 12 = -10 / 12 = -5/6
Step 4: Determine the values of x using inverse trigonometric functions.
Now, we need to solve for x using the values of t.
For t = 1:
Since we are given the interval (∏/2, ∏), x is between ∏/2 and ∏. In this interval, cos(x) is negative. So, we need to find the arccosine of 1 that falls in this interval.
Using inverse trigonometric functions, x = arccos(1).
Since the arccosine of 1 is 0, x = 0.
For t = -5/6:
Again, we need to find the arccosine of -5/6 in the interval (∏/2, ∏). In this interval, cos(x) is negative.
Using inverse trigonometric functions, x = arccos(-5/6).
This cannot be expressed in terms of elementary inverse trigonometric functions.
Therefore, the solution for x in terms of inverse trigonometric functions is:
x = 0