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One pivotal concept in trigonometry is the double-angle identity. This identity provides a way to express trigonometric functions of double angles in terms of single angles. For sine, the double-angle identity is: \[ \sin(2x) = 2\sin(x)\cos(x) \] This formula is particularly helpful because it transforms complex trigonometric equations into more manageable forms, easing the process of solving.When dealing with equations like \( \sin(2x) = \sin(x) \), employing this identity simplifies things significantly. By substituting \( \sin(2x) \) with \( 2\sin(x)\cos(x) \), you create a new equation: \[ 2\sin(x)\cos(x) - \sin(x) = 0 \] In essence, the double-angle identity directly connects two identities, allowing you to move from two intertwined angles to more simple expressions. This makes solving such equations much easier!

Once you've applied an identity and simplified an equation, the next step is often to factor it. Factoring is the process of rewriting the equation in terms of products. For the equation \( 2\sin(x)\cos(x) - \sin(x) = 0 \), factoring involves pulling out common factors. Here, \( \sin(x) \) is common in both terms. By factoring \( \sin(x) \) out, we get:\[ \sin(x)(2\cos(x) - 1) = 0 \] This transformation makes it easier to solve the equation because once it's factored, you can use the zero-product property. The zero-product property states that if the product of two numbers is zero, at least one of the numbers must be zero.So, either \( \sin(x) = 0 \) or \( 2\cos(x) - 1 = 0 \). Factoring simplifies equations to basic components that are much easier to solve for solutions.

After simplifying and factoring a trigonometric equation, the following task is to solve it. Solving involves determining the values of the variable that satisfy the equation. In our factored equation \( \sin(x)(2\cos(x) - 1) = 0 \), we solve by setting each factor equal to zero:- Solving \( \sin(x) = 0 \), we find solutions: - \( x = 0 \) - \( x = \pi \)- Solving \( 2\cos(x) - 1 = 0 \), we find: - \( 2\cos(x) = 1 \) - \( \cos(x) = \frac{1}{2} \) - Solutions are \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \)These solutions must be verified as valid within the given interval \([0, 2\pi)\). This process not only demands finding values but ensures they are appropriate within the context or boundaries provided, making sure we address the problem correctly.