solve for x sin4x+sin2x-sin6x=0

To solve the equation sin(4x) + sin(2x) - sin(6x) = 0, we can start by utilizing trigonometric identities to simplify the equation. One effective approach is to express the sine functions in terms of a common angle or to use sum-to-product identities.

Rearranging the Equation

Let’s first rewrite the equation for clarity:

sin(4x) + sin(2x) = sin(6x)

Using the Sum-to-Product Identities

We can use the sum-to-product identities for sine, which state:

• sin(a) + sin(b) = 2 sin((a + b)/2) cos((a - b)/2)

Applying this identity to sin(4x) + sin(2x), we have:

• a = 4x, b = 2x
• sin(4x) + sin(2x) = 2 sin((4x + 2x)/2) cos((4x - 2x)/2)
• sin(4x) + sin(2x) = 2 sin(3x) cos(x)

Substituting Back into the Equation

Now we can substitute this back into our rearranged equation:

2 sin(3x) cos(x) = sin(6x)

Expressing sin(6x)

Next, we can express sin(6x) using the double angle identity:

• sin(6x) = 2 sin(3x) cos(3x)

Setting the Equation

Now we can rewrite our equation:

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

Dividing both sides by 2 (assuming sin(3x) ≠ 0):

sin(3x) cos(x) = sin(3x) cos(3x)

Factoring Out sin(3x)

This gives us two cases to consider:

• Case 1: sin(3x) = 0
• Case 2: cos(x) = cos(3x)

Solving Case 1

For the first case, sin(3x) = 0, we find:

3x = nπ, where n is an integer.

Thus, x = nπ/3.

Solving Case 2

In the second case, cos(x) = cos(3x). This leads us to the general solutions:

• x = 3x + 2kπ or x = -3x + 2kπ, where k is an integer.

Simplifying the Cases

From the first equation:

2x = 2kπ, leading to x = kπ.

From the second equation:

4x = 2kπ, leading to x = kπ/2.

Summary of Solutions

Combining both cases, we find that:

• x = nπ/3 (from Case 1)
• x = kπ (from Case 2)
• x = kπ/2 (from Case 2)

In conclusion, the solutions to the equation sin(4x) + sin(2x) - sin(6x) = 0 are given by the values of x that can be represented as nπ/3, kπ, or kπ/2 for integers n and k. This provides a comprehensive set of angles at which the original equation holds true.

2cos3xsinx + 2sinxcox=0

=>sinx=0

=>x= npi...................... n is 1,2,3............

Also,

2cos3x + cosx=0

8cos³x-5cosx=0

cosx=0

=> x is (2k+1)pi/2......................................................................................

Also,

8cos²x-5=0

cosx=+-(5/8)^1/2.................................................

All the above are solutions for the given equation

hi prajwal

you made a slight mistake