Question

1)cos x cos2x = sin x sin2x    2)2cos^2 2x+3cos^2x=2   3)(sinx+cosx)^2=1+cosx 

Answer 1

Equation 1: cos(x) + cos(2x) = sin(x) + sin(2x)

To solve this equation, we can start by simplifying the trigonometric expressions on both sides.

Let's express sin(2x) and cos(2x) in terms of sin(x) and cos(x) using the double angle identities:

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

Now we can substitute these expressions into the equation:

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

Expanding the equation further:

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

Rearranging the terms:

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

Using the identity cos^2(x) - sin^2(x) = cos(2x), the equation becomes:

cos(2x) - 2sin(x)cos(x) + cos(x) - sin(x) = 0

Now, let's combine like terms:

cos(2x) + (cos(x) - sin(x))(1 - 2sin(x)) = 0

To solve this equation, we need to consider two cases: when (1 - 2sin(x)) = 0 and when cos(2x) = 0.

Case 1: (1 - 2sin(x)) = 0 Solving this equation for sin(x), we get sin(x) = 1/2. This gives us two solutions: x = π/6 and x = 5π/6.

Case 2: cos(2x) = 0 Solving this equation for x, we get x = π/4 and x = 3π/4.

Therefore, the solutions to the equation cos(x) + cos(2x) = sin(x) + sin(2x) are x = π/6, x = 5π/6, x = π/4, and x = 3π/4.

Equation 2: 2cos^2(2x) + 3cos^2(x) = 2 + 3

To solve this equation, let's simplify the trigonometric expressions first.

Using the identity cos(2x) = 2cos^2(x) - 1, the equation becomes:

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

Expanding and simplifying:

4cos^2(x) - 2 + 3cos^2(x) = 5

Combining like terms:

7cos^2(x) - 2 = 5

Adding 2 to both sides:

7cos^2(x) = 7

Dividing both sides by 7:

cos^2(x) = 1

Taking the square root of both sides:

cos(x) = ±1

The cosine function takes on values between -1 and 1, so the only solutions are cos(x) = 1 and cos(x) = -1.

For cos(x) = 1, x = 0 + 2πn (where n is an integer). For cos(x) = -1, x = π + 2πn (where n is an integer).

Therefore, the solutions to the equation 2cos^2(2x) + 3cos^2(x) = 2 + 3 are x = 0 + 2πn and x = π + 2πn (where n is an integer).

Equation 3: (sin(x) + cos(x))^2 = 1 + cos(x)

To solve this equation, let's expand the square on the left side and simplify:

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

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, the equation becomes:

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

Subtracting 1 and cos(x) from both sides:

2sin(x)cos(x) - cos(x) = 0

Factoring out cos(x):

cos(x)(2sin(x) - 1) = 0

This equation can be satisfied if either cos(x) = 0 or (2sin(x) - 1) = 0.

For cos(x) = 0, x = π/2 + πn (where n is an integer).

For 2sin(x) - 1 = 0, sin(x) = 1/2. This gives us two solutions: x = π/6 and x = 5π/6.

Therefore, the solutions to the equation (sin(x) + cos(x))^2 = 1 + cos(x) are x = π/2 + πn, x = π/6, and x = 5π/6 (where n is an integer).