Question

If \(\sin 5 x+\sin 3 x+\sin x=0\), then the value of \(x\) other than zero in the interval \(0

Short answer

Question: Determine if there is a nonzero value for x in the interval \(0 < x \leq \frac{\pi}{2}\) that satisfies the given equation: \(\sin 5 x+\sin 3 x+\sin x= 0\). Answer: There is no nonzero value for \(x\) in the interval \(0 < x \leq \frac{\pi}{2}\) that satisfies the given equation.

Step-by-step solution

Apply the sine addition formula

Let's first apply the sine addition formula on the first two terms: \(\sin p + \sin q = 2 \sin (\frac{p+q}{2}) \cos (\frac{p-q}{2})\) Here, we have \(p = 5x\) and \(q = 3x\). Applying the formula, we get: \(2 \sin(\frac{5x + 3x}{2}) \cos(\frac{5x - 3x}{2}) + \sin x = 0\)

Simplify the expression

Now, let's simplify the expression: \(2 \sin(4x) \cos(x) + \sin x = 0\)

Factor out common term

We can factor out the common term \(\sin x\): \(\sin x (2 \cos x \sin 4x + 1) = 0\)

Set each factor to zero and solve for x

Now, either \(\sin x = 0\) or \((2 \cos x \sin 4x + 1) = 0\). Since we are looking for a nonzero value of \(x\) in the interval \(0 < x \leq \frac{\pi}{2}\), we can ignore the first case, sin(x)=0. Now, let's focus on the second factor and solve for \(x\): \(2 \cos x \sin 4x + 1 = 0\)

Use sin(2x) formula to simplify

Let's apply the sine double-angle formula: \(\sin 2x = 2 \sin x \cos x\): \(4 \sin(2x) \sin x + 1 = 0\) From here, we use the sine double-angle formula again for \(4\sin(2x)\sin(x)\) \(4 \sin x (2 \cos x \sin x) + 1 = 0\)

Simplify and solve for x

Now, we have: \(\sin x (4 \cos x \sin x + 1) = 0\) This expression is the same as the one in step 3, and the first factor is repeating while the second factor has no solution. Hence, there is no other solution for \(x\). In conclusion, no value for \(x\) other than zero in the interval \(0<x\leq \frac{\pi}{2}\) satisfies the given equation.

Key concepts

Sine Addition Formula

The sine addition formula is pivotal for solving trigonometric equations involving sums of sine terms. It helps simplify expressions and makes solving complex equations more manageable. The formula is expressed as:\[ \sin p + \sin q = 2 \sin \left(\frac{p+q}{2}\right) \cos \left(\frac{p-q}{2}\right) \]- **Application**: It is used to combine two sine terms into a single expression, which is particularly helpful when dealing with multiple-angle sines. - **Benefit**: This simplification can reveal patterns or factors that make further solving steps straightforward. For example, in the problem, by setting \(p = 5x\) and \(q = 3x\), the expression is transformed, simplifying the complexity and opening the path to further factorization.

Double Angle Formula

The double angle formula is an incredibly useful tool in trigonometry for simplifying expressions where angles are doubled. Specifically, the double angle formula for sine is:\[ \sin 2x = 2 \sin x \cos x \]- **Purpose**: This allows one to express trigonometric functions of double angles in terms of single angles. - **Example Usage**: In the given problem, when simplifying the original equation to \(2 \cos x \sin 4x + 1 = 0\), identifying terms of the form \(\sin 2x\) can help transform them using this formula. The benefit of using such formulas is a streamlined expression, which is easier to factor and solve.

Trigonometric Identities

Trigonometric identities are foundational equations involving trigonometric functions that hold true for all angle measures. These identities offer a toolbox for simplifying complex expressions. Some of the most commonly used include:- **Pythagorean Identity**: \( \sin^2 x + \cos^2 x = 1 \) - **Sum to Product**: Useful for transforming sums into products, helping with simplification or factorization in equations. When dealing with the problem at hand, such identities are applied to bridge between forms and finding solutions. Repeated use of specific identities like the double-angle formula assists in breaking down intricate trigonometric equations into solvable components, maintaining an efficient route to handle alternating sine and cosine terms.