prove that the general solution of the equation

( cos 3x – sin 3x)² = sin 2x + 1 is ( 2n + 1)* pi/4

To prove that the general solution of the equation \( ( \cos 3x - \sin 3x )^2 = \sin 2x + 1 \) is \( x = \frac{(2n + 1) \pi}{4} \), where \( n \) is any integer, we can approach this step-by-step, breaking down the equation and analyzing it systematically.

Step 1: Simplifying the Left Side

First, let's simplify the left side of the equation:

• We have \( ( \cos 3x - \sin 3x )^2 \).
• Using the identity \( (a - b)^2 = a^2 - 2ab + b^2 \), we can expand this:
• Thus, we get \( \cos^2 3x - 2 \cos 3x \sin 3x + \sin^2 3x \).

Knowing that \( \cos^2 \theta + \sin^2 \theta = 1 \), we can further simplify:

So, the left side becomes \( 1 - 2 \cos 3x \sin 3x \), which can be rewritten using the double angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):

Therefore, \( 1 - \sin 6x \).

Step 2: Analyzing the Right Side

Now, let's examine the right side of the equation:

• We have \( \sin 2x + 1 \).
• We know that the maximum value of \( \sin 2x \) is 1, which means the right side ranges from 1 to 2.

Step 3: Setting the Equation

Now, equating both sides gives us:

\( 1 - \sin 6x = \sin 2x + 1 \).

By rearranging, we can express this as:

\( -\sin 6x = \sin 2x \).

This leads to:

\( \sin 6x + \sin 2x = 0 \).

Step 4: Utilizing the Sum-to-Product Identity

To solve the equation \( \sin 6x + \sin 2x = 0 \), we can apply the sum-to-product identities:

This states that \( a + b = 0 \) leads us to \( \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) = 0 \).

Setting \( A = 6x \) and \( B = 2x \), we find:

• \( 6x + 2x = 8x \) which leads to \( \sin 8x = 0 \),
• and \( 6x - 2x = 4x \) which leads to \( \cos 4x = 0 \).

Step 5: Solving for x

From \( \sin 8x = 0 \), we derive solutions:

\( 8x = n\pi \Rightarrow x = \frac{n\pi}{8} \) for integers \( n \).

From \( \cos 4x = 0 \), we derive solutions:

\( 4x = \frac{\pi}{2} + m\pi \Rightarrow x = \frac{(2m + 1)\pi}{8} \) for integers \( m \).

Combining Solutions

To find a common form, we can express both solutions in terms of \( n \). Setting \( n = 2m + 1 \), we arrive at:

\( x = \frac{(2n + 1)\pi}{4} \) where \( n \) is an integer.

Conclusion

Thus, we conclude that the general solution to the equation \( ( \cos 3x - \sin 3x )^2 = \sin 2x + 1 \) is indeed \( x = \frac{(2n + 1)\pi}{4} \) for any integer \( n \). This comprehensive breakdown illustrates the logical steps taken to arrive at the solution, ensuring clarity throughout the process.