Question

The values of x in $0\le x\le\pi$ such that $\cos 2x=\cos x$ are

• A. $0$ and $\frac{2\pi}{3}$
• B. $\frac{\pi}{3}$ and $\frac{2\pi}{3}$
• C. $0$ and $\frac{\pi}{3}$
• D. $\frac{\pi}{4}$ and $\frac{\pi}{3}$
• E. $0$ and $\frac{\pi}{2}$

Hint:

General Solution Tip: When $\cos A = \cos B$, the general solution is $A = 2n\pi \pm B$. Here, $2x = 2n\pi \pm x$, which gives $x = 2n\pi$ or $3x = 2n\pi$. Testing integers for $n$ quickly gives $0$ and $2\pi/3$ in the domain $[0, \pi]$.

Answer 1

The Correct Option is A

Concept:
To solve a trigonometric equation with multiple angles (like $2x$ and $x$), use double angle identities to unify the terms. The identity $\cos(2x) = 2\cos^2x - 1$ transforms the equation into a standard quadratic equation in terms of $\cos x$.

Step 1: State the equation and apply the identity.
Given: $$\cos 2x = \cos x$$ Substitute $\cos 2x$ with $2\cos^2x - 1$: $$2\cos^2x - 1 = \cos x$$

Step 2: Form a standard quadratic equation.
Move all terms to one side to set the equation to zero: $$2\cos^2x - \cos x - 1 = 0$$

Step 3: Factor the quadratic equation.
Let $u = \cos x$. The equation is $2u^2 - u - 1 = 0$. Factor by splitting the middle term: $$2u^2 - 2u + u - 1 = 0$$ $$2u(u - 1) + 1(u - 1) = 0$$ $$(2u + 1)(u - 1) = 0$$ Substituting $\cos x$ back in: $$(2\cos x + 1)(\cos x - 1) = 0$$

Step 4: Solve for the possible values of cosine.
Set each factor to zero: Case 1: $\cos x - 1 = 0 \implies \cos x = 1$ Case 2: $2\cos x + 1 = 0 \implies \cos x = -\frac{1}{2}$

Step 5: Find the corresponding angles in the given domain.
We must find $x$ within the restricted interval $0 \le x \le \pi$. For Case 1 ($\cos x = 1$): The only angle in this domain is $x = 0$. For Case 2 ($\cos x = -\frac{1}{2}$): Cosine is negative in the second quadrant. The reference angle is $\frac{\pi}{3}$, so $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$. The solutions are $0$ and $\frac{2\pi}{3}$. Hence the correct answer is (A) $0$ and $\frac{2\pi{3}$}.