If 1-sin2x=cosx-sinx then find the general form of x

To solve the equation \(1 - \sin(2x) = \cos(x) - \sin(x)\), we will start by using some trigonometric identities and algebraic manipulation. Let's break down the equation step by step.

Step 1: Use Trigonometric Identities

Recall the double angle identity for sine: \(\sin(2x) = 2\sin(x)\cos(x)\). We can substitute this into our equation:

\[1 - 2\sin(x)\cos(x) = \cos(x) - \sin(x)\]

Step 2: Rearranging the Equation

Next, let's rearrange the equation to bring all terms to one side:

\[1 - \cos(x) + \sin(x) - 2\sin(x)\cos(x) = 0\]

Step 3: Grouping Terms

Now, let's rearrange and group the terms appropriately:

\[1 - \cos(x) + \sin(x) - 2\sin(x)\cos(x) = 0\]

Notice that we have both sine and cosine terms. This suggests that we can potentially factor the equation or use substitution.

Step 4: Substituting Values

Instead of trying to factor directly, let's isolate \(\sin(x)\) and \(\cos(x)\). We can rearrange the equation as follows:

\[1 - \cos(x) = 2\sin(x)\cos(x) - \sin(x)\]

Factoring \(\sin(x)\) from the right side gives us:

\[1 - \cos(x) = \sin(x)(2\cos(x) - 1)\]

Step 5: Solving for Specific Cases

Now we can consider two cases based on the factorization:

• Case 1: \(1 - \cos(x) = 0\)
• Case 2: \(\sin(x) = 0\)

Case 1: \(1 - \cos(x) = 0\)

This simplifies to \(\cos(x) = 1\). The general solution for \(\cos(x) = 1\) is:

\[x = 2n\pi \quad (n \in \mathbb{Z})\]

Case 2: \(\sin(x) = 0\)

The solutions for \(\sin(x) = 0\) occur at:

\[x = n\pi \quad (n \in \mathbb{Z})\]

Combining Solutions

Thus, the general solutions to the original equation are:

\[x = n\pi \quad \text{and} \quad x = 2m\pi \quad (n, m \in \mathbb{Z})\]

Both sets of solutions include all integer multiples of \(\pi\), so we can summarize that the general form of \(x\) that satisfies the equation is:

\[x = n\pi \quad (n \in \mathbb{Z})\]

This encapsulates both cases since \(n\) can be odd or even, covering all potential solutions. If you have any further questions or need clarification on any part of the process, feel free to ask!