Solve for x on the interval [0,π/2):
cos^3(2x) + 3cos^2(2x) + 3cos(2x) = 4
I havent done trig for a while so what exactly does that mean in solving for x on that interval and how would i go about doing that?
Where are you getting these tough trig questions??
how about letting cos 2x = t
then your equation becomes
t^3 + 3t^2 + 3t - 4 = 0
using Newton's Method I was able to come up with ONE real root , t=.709976
so cos 2x = .709976
for 2x= .7813322 radians
and finally by dividing by 2
x = appr. .39
You can factorize the cubic equation Reiny gave above:
t^3 + 3t^2 + 3t - 4 = 0 --->
(t+1)^3 = 5 --->
t = -1 + 5^(1/3)
Count, now that was clever.
good for you to recognize the sequence
1,3,3,(1) in Pascal's triangle.
To solve for x on the interval [0,π/2), we start by simplifying the given equation by letting cos(2x) = t. This substitution allows us to transform the equation into a simpler form: t^3 + 3t^2 + 3t - 4 = 0.
Next, we can use numerical methods, such as Newton's Method, to find the value of t that satisfies the equation. In this case, you have already found that t = 0.709976.
Now that we have the value of t, we can solve for x. Since cos(2x) = t, we can write 2x = 0.7813322 radians by taking the inverse cosine of t. Finally, by dividing by 2, we find x ≈ 0.39 as the solution.
Alternatively, you can factorize the cubic equation by recognizing the pattern in Pascal's triangle:
(t+1)^3 = 5
Taking the cube root of both sides, we get:
t = -1 + 5^(1/3)
This gives us the same value for t, which we can then use to solve for x as described earlier.
As for where these questions come from, they are designed to challenge your trigonometry skills and apply various problem-solving techniques. They may be sourced from textbooks, practice exams, or real-world problems.