Problem 26
Question
Solve: \(\cos x \cos 2 x \cos 4 x=\frac{1}{4}\) \(0 \leq x \leq \pi\)
Step-by-Step Solution
Verified Answer
The solutions to the equation \(\cos x \cos 2x \cos 4x = \frac{1}{4}\) in the interval \(0 ≤ x ≤ \pi\) are \(x = \frac{\pi}{21}, \frac{2 \pi}{21}, \frac{8\pi}{21},\) and \(\frac{10\pi}{21}.\)
1Step 1: Use of Product-to-Sum Identity
Rewrite the equation \(\cos x \cos 2x \cos 4x = \frac{1}{4}\) using product-to-sum identities. The identity \(\cos a \cos b = \frac{1}{2} [\cos(a+b)+\cos(a-b)]\) can be used twice. Firstly applying it to \(\cos x \cos 2x\), then to \(\cos(3x)\cos(4x)\).
2Step 2: Further Simplification
Keep simplifying the equation. Multiply it by 2 to get \(2 \cos x \cos 2x \cos 4x = \frac{1}{2}\), which can be rewritten as \(\cos 7x = \frac{1}{2}\). This simple equation can be solved using known values of cosine.
3Step 3: Solve for x
Finally, we can use the fact that \(\cos(\pi/3) = \frac{1}{2}\) and that cosine is also positive in the fourth quadrant (\(2 \pi - \pi/3\)). Hence the solutions are all \(x = \frac{\pi}{21}\) (from first solution) and \(x = \frac{2 \pi}{21}\) (from second solution) plus any multiples of \(2\pi/7\) (from periodicity of cosine), which lie in the desired interval \(0 ≤ x ≤ \pi\).
Key Concepts
Product-to-Sum IdentityCosine Equation SolutionsPeriodicity of Trigonometric Functions
Product-to-Sum Identity
Trigonometric identities are tools that allow us to simplify complex expressions. Among these, the product-to-sum identities convert products of trigonometric functions into sums. This can make equations much easier to handle. Specifically, the identity for cosine is:
This approach greatly simplifies the equation, often reducing it to a manageable form that can be solved algebraically.
- \( \cos a \cos b = \frac{1}{2} [\cos(a+b) + \cos(a-b)] \)
- First, convert \( \cos x \cos 2x \) into a sum using the identity.
- Then, in the resulting expression, apply the identity again to combine terms that include \( \cos 4x \).
This approach greatly simplifies the equation, often reducing it to a manageable form that can be solved algebraically.
Cosine Equation Solutions
Solving trigonometric equations like \( \cos 7x = \frac{1}{2} \) involves understanding when the cosine function equals specific values. The cosine function is known to equal \( \frac{1}{2} \) at certain key angles:
- The first is \( x = \frac{\pi}{3} \).
- In the unit circle, cosine is also positive in the fourth quadrant, meaning \( \cos(2\pi - \frac{\pi}{3}) = \frac{1}{2} \), leading to solutions around \( \frac{5\pi}{3} \).
- We solve for \( x \) by dividing these known angles by 7, such as \( x = \frac{\pi}{21} \).
- Solutions repeat periodically; for \( \cos 7x \), periodicity results in expressions like \( x = \frac{\pi}{21} + \frac{2k\pi}{7} \) for integer \( k \).
Periodicity of Trigonometric Functions
Periodicity is a core property of trigonometric functions, meaning these functions repeat their values in regular intervals. This characteristic is crucial for solving equations as it gives us an infinite set of solutions.For cosine, the function repeats every \( 2\pi \) radians. However, when dealing with transformations, the apparent period might differ. In the equation \( \cos 7x \) = \( \frac{1}{2} \), the factor 7 affects the period:
- The new period is \( \frac{2\pi}{7} \).
- This means every solution should account for this change, repeating every \( \frac{2\pi}{7} \) units.
Other exercises in this chapter
Problem 25
Solve: \(\sin 2 x+\sin x+\cos 2 x+\cos x+1=0\)
View solution Problem 26
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View solution Problem 27
Solve the following equations and tick the correct one. The maximum value of \(5 \sin \theta+3 \sin (\theta-\alpha)\) is 7 , then the set of all possible values
View solution Problem 27
Solve: \(\sin 3 \alpha=4 \sin \alpha \cdot \sin (x+\alpha) \cdot \sin (x-\alpha)\)
View solution