Solve the equation on the interval 0 less than or equals theta less than 2 pi 0≤θ<2π. 2cos^(2)\theta +\sqrt(2cos\theta )=2cos\theta +\sqrt(2)
why do you write "the interval 0 less than or equals theta less than 2 pi" and then go right ahead and say "0≤θ<2π" ? Just like to repeat yourself?
2cos^2θ + √(2cosθ) = 2cosθ + √2
Just by inspection, θ=0
and θ cannot be in QII or QIII, since there cosθ < 0
and since the function is even, solutions in QIV mirror those in QI.
To solve the equation 2cos^2θ + √(2cosθ) = 2cosθ + √2 on the interval 0≤θ<2π, we can follow these steps:
Step 1: Let's rewrite the equation in a more manageable form. Since both sides contain a square root term, we'll isolate the square root first:
√(2cosθ) = 2cosθ + √2 - 2cos^2θ
Step 2: Now, let's square both sides of the equation to eliminate the square root:
(√(2cosθ))^2 = (2cosθ + √2 - 2cos^2θ)^2
Simplifying gives us:
2cosθ = (2cosθ + √2 - 2cos^2θ)^2
Step 3: Expand and simplify the right side of the equation:
2cosθ = (2cosθ + √2 - 2cos^2θ)(2cosθ + √2 - 2cos^2θ)
2cosθ = 4cos^2θ + 2√2cosθ - 4cos^4θ + 2√2cosθ - √2cosθ - 2√2cos^3θ + 4cos^3θ - 2cosθ
2cosθ = 4cos^2θ - 4cos^4θ - √2cosθ - 2√2cos^3θ
Step 4: Rearrange the equation and simplify the terms:
4cos^4θ - 4cos^2θ - √2cosθ - 2√2cos^3θ + 2cosθ = 0
4cos^4θ - 2cos^2θ - √2cosθ - 2√2cos^3θ = 0
Step 5: Factor out common terms:
2cosθ(2cos^3θ - cosθ - √2 - √2cos^2θ) = 0
Step 6: We have two potential solutions for the equation to hold:
a) cosθ = 0
b) 2cos^3θ - cosθ - √2 - √2cos^2θ = 0
Solving for each solution:
a) For cosθ = 0, the solutions on the given interval are θ = π/2 and θ = 3π/2.
b) To solve 2cos^3θ - cosθ - √2 - √2cos^2θ = 0, we can use numerical methods or calculators as it doesn't have simple closed-form solutions.