Question

Sin^2x+sin^22x+cos^23x+cos^24x=2

Answer 1

Sin²x+sin²2x+cos²3x+cos²4x=2
используем формулы  sin²x=(1-cos2x)/2    cos²x=(cos2x+1)/2
(1-cos2x)/2+(1-cos4x)/2+(cos6x+1)/2+(1+cos8x)/2=2
1-cos2x+1-cos4x+cos6x+1+1+cos8x=4
-cos2x-cos4x+cos6x+cos8x=0
(cos6x+cos8x)-(cos2x+cos4x)=0
используем формулу  cosα+cosβ=2cos(α+β)/2cos(α-β)/2
2cos7xcosx-2cos3xcosx-0
2cosx(cos7x-cos3x)=0
cosx=0  x=π/2+πn  n∈Z
cos7x-cos3x=0
-2sin2xsin5x=0
sin2x=0
2x=πk  x=πk/2  k∈Z
sin5x=0
5x=πm  x=πm/5  m∈Z

Answer 2

To solve the trigonometric equation sin^2(x) + sin^2(2x) + cos^2(3x) + cos^2(4x) = 2, we need to find the values of x that satisfy the equation.

Let's simplify the equation step by step:

1. Using the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we can simplify the first two terms: sin^2(x) + sin^2(2x) = 1 + sin^2(2x)

2. Similarly, using the same identity for the last two terms: cos^2(3x) + cos^2(4x) = 1 + cos^2(4x)

Now, the equation becomes:

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

Simplify further:

sin^2(2x) + cos^2(4x) = 0

Now, we use another trigonometric identity sin^2(θ) + cos^2(θ) = 1, but in this case, the sum of the squares of sine and cosine is equal to 0, which means both terms must individually be equal to 0:

sin^2(2x) = 0 cos^2(4x) = 0

1. Solving sin^2(2x) = 0:

sin^2(2x) = 0 sin(2x) = 0

To find the solutions for sin(2x) = 0, we look for the values of 2x for which the sine function is 0. The sine function is 0 at multiples of π (pi), so:

2x = nπ x = nπ/2

where n is an integer.

2. Solving cos^2(4x) = 0:

cos^2(4x) = 0 cos(4x) = 0

To find the solutions for cos(4x) = 0, we look for the values of 4x for which the cosine function is 0. The cosine function is 0 at odd multiples of π/2:

4x = (2n + 1)π/2 x = (2n + 1)π/8

where n is an integer.

So, the solutions for the original equation sin^2(x) + sin^2(2x) + cos^2(3x) + cos^2(4x) = 2 are:

x = nπ/2 or x = (2n + 1)π/8, where n is an integer.