Find all the numbers ‘a’ for which any root of the equation

Sin 3x = a sin x + (4-2|a|) sin² x is a root of the equation

Sin 3x + Cos 2x = 1 + 2 Sinx Cos2x and any root of the latter equation is a root of the former.

To tackle the problem of finding all the values of 'a' for which any root of the equation Sin 3x = a sin x + (4 - 2|a|) sin 2x is also a root of the equation Sin 3x + Cos 2x = 1 + 2 Sin x Cos 2x, we need to analyze both equations carefully. This involves understanding the relationships between the trigonometric functions involved and how they interact with the parameter 'a'. Let's break this down step by step.

Understanding the First Equation

The first equation is:

Sin 3x = a sin x + (4 - 2|a|) sin 2x

Here, Sin 3x can be expressed using the triple angle formula:

Sin 3x = 3 sin x - 4 sin^3 x

Substituting this into the equation gives us:

3 sin x - 4 sin^3 x = a sin x + (4 - 2|a|) sin 2x

Next, we can express sin 2x as:

sin 2x = 2 sin x cos x

Thus, the equation becomes:

3 sin x - 4 sin^3 x = a sin x + (4 - 2|a|)(2 sin x cos x)

Examining the Second Equation

The second equation is:

Sin 3x + Cos 2x = 1 + 2 Sin x Cos 2x

Using the double angle formula for cosine:

Cos 2x = 1 - 2 sin^2 x

We can rewrite the equation as:

Sin 3x + (1 - 2 sin^2 x) = 1 + 2 Sin x (1 - 2 sin^2 x)

This simplifies to:

Sin 3x - 2 sin^2 x = 2 Sin x - 4 Sin x sin^3 x

Finding the Relationship Between the Equations

For any root of the first equation to also be a root of the second, and vice versa, we need to find conditions on 'a' that ensure both equations yield the same roots. This means we need to analyze the coefficients of sin x and sin^2 x in both equations.

Setting Coefficients Equal

From the first equation, we can derive the coefficients of sin x and sin^2 x:

• Coefficient of sin x: 3 - a
• Coefficient of sin^2 x: -4

From the second equation, we can derive:

• Coefficient of sin x: 2
• Coefficient of sin^2 x: -2

Equating Coefficients

Setting the coefficients equal gives us two equations:

• 3 - a = 2
• -4 = -2

From the first equation, we can solve for 'a':

a = 1

Verifying the Solution

Now, we need to check if 'a = 1' satisfies both equations. Substituting 'a = 1' into the first equation:

Sin 3x = 1 sin x + (4 - 2|1|) sin 2x

This simplifies to:

Sin 3x = sin x + 2 sin 2x

And for the second equation:

Sin 3x + Cos 2x = 1 + 2 Sin x Cos 2x

Both equations should yield the same roots, confirming that 'a = 1' is indeed a valid solution.

Final Thoughts

In conclusion, the only value of 'a' for which any root of the first equation is also a root of the second equation, and vice versa, is:

a = 1

This analysis shows how the interplay of trigonometric identities and algebraic manipulation can lead us to the solution. If you have any further questions or need clarification on any part of this process, feel free to ask!