Question

Sin^(2)x+sin^(2)2x-sin^(2)3x=0

Answer 1

1-cos2x+1-cos4x-1+cos6x=0
2 cos^(2)3x-2cos3xcosx=0
Cos3x=0 или сos3x=Cosx, откуда

Answer 2

To solve the equation sin^(2)x + sin^(2)2x - sin^(2)3x = 0, we can simplify it by using the trigonometric identity:

sin^(2)(2x) = (1/2)(1 - cos(4x))

sin^(2)(3x) = (1/2)(1 - cos(6x))

Substituting these values into the equation, we have:

sin^(2)x + (1/2)(1 - cos(4x)) - (1/2)(1 - cos(6x)) = 0

Now, let's simplify further:

sin^(2)x + (1/2) - (1/2)cos(4x) - (1/2) + (1/2)cos(6x) = 0

Rearranging the terms, we get:

sin^(2)x - (1/2)cos(4x) + (1/2)cos(6x) = 0

Using the trigonometric identity:

sin^(2)x = (1/2)(1 - cos(2x))

We can substitute this back into the equation:

(1/2)(1 - cos(2x)) - (1/2)cos(4x) + (1/2)cos(6x) = 0

Simplifying further:

(1/2) - (1/2)cos(2x) - (1/2)cos(4x) + (1/2)cos(6x) = 0

Now, we have an equation involving cosines. To solve it, we'll need to analyze the properties of cosine functions or use numerical methods to find the solutions. Unfortunately, it is not possible to provide a general analytical solution for this equation due to the complex nature of trigonometric functions.