Question

Sinx+2sin2x+sin3x=|1+2cosx+cos2x|

Answer 1

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Answer 2

To solve the trigonometric equation sin(x) + 2sin(2x) + sin(3x) = |1 + 2cos(x) + cos(2x)|, we need to analyze it step by step.

1. Let's simplify the right-hand side (RHS) of the equation first:

   |1 + 2cos(x) + cos(2x)|

2. Now, let's look at the left-hand side (LHS) of the equation:

   sin(x) + 2sin(2x) + sin(3x)

3. We know that the modulus (absolute value) of a quantity is always positive or zero. So, to eliminate the absolute value, we can consider two separate cases:

   Case 1: 1 + 2cos(x) + cos(2x) is positive. Case 2: 1 + 2cos(x) + cos(2x) is negative or zero.

Let's start with Case 1:

Case 1: 1 + 2cos(x) + cos(2x) > 0

In this case, we can remove the absolute value:

1 + 2cos(x) + cos(2x) = 1 + 2cos(x) + cos(2x)

Now, the equation becomes:

sin(x) + 2sin(2x) + sin(3x) = 1 + 2cos(x) + cos(2x)

Next, we can use trigonometric identities to express sin(2x) and cos(2x) in terms of sin(x) and cos(x):

sin(2x) = 2sin(x)cos(x) cos(2x) = cos^2(x) - sin^2(x)

Substitute these into the equation:

sin(x) + 4sin(x)cos(x) + sin(3x) = 1 + 2cos(x) + (cos^2(x) - sin^2(x))

Now, let's simplify further:

sin(x) + 4sin(x)cos(x) + sin(3x) = 1 + 2cos(x) + cos^2(x) - sin^2(x)

Now, we have an equation involving multiple trigonometric functions. To proceed, we need to use additional trigonometric identities and algebraic manipulations.

Case 2: 1 + 2cos(x) + cos(2x) ≤ 0

In this case, we can also remove the absolute value but with a negative sign:

-(1 + 2cos(x) + cos(2x)) = 1 + 2cos(x) + cos(2x)

Now, the equation becomes:

sin(x) + 2sin(2x) + sin(3x) = -(1 + 2cos(x) + cos(2x))

Now, just like in Case 1, we'll use trigonometric identities to express sin(2x) and cos(2x) in terms of sin(x) and cos(x):

sin(2x) = 2sin(x)cos(x) cos(2x) = cos^2(x) - sin^2(x)

Substitute these into the equation:

sin(x) + 4sin(x)cos(x) + sin(3x) = -(1 + 2cos(x) + (cos^2(x) - sin^2(x)))

Now, simplify further:

sin(x) + 4sin(x)cos(x) + sin(3x) = -(1 + 2cos(x) + cos^2(x) - sin^2(x))

Again, we have an equation involving multiple trigonometric functions. To find the solutions for the given equation, we'd need to proceed with more trigonometric identities and algebraic manipulations. Unfortunately, the equation is quite complex, and the solutions may not be straightforward to find.

Keep in mind that trigonometric equations can have multiple solutions, and finding exact solutions for this equation might be challenging. You may use numerical methods or graphing techniques to approximate the solutions in both cases.