Question

3sin^2(x)+5cos^2(x)-2cos(2x)+4sin2x=0

Answer 1

Решение дано на фото.

Answer 2

To solve the trigonometric equation 3sin^2(x) + 5cos^2(x) - 2cos(2x) + 4sin(2x) = 0, we'll use trigonometric identities and algebraic manipulation to simplify and find the solutions.

Step 1: Apply double angle and Pythagorean identities Recall the double angle identities:

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

Also, the Pythagorean identity:

• sin^2(x) + cos^2(x) = 1

Step 2: Substitute the identities into the equation Replace cos(2x) and sin(2x) using the double angle identities:

3sin^2(x) + 5cos^2(x) - 2(cos^2(x) - sin^2(x)) + 4(2sin(x)cos(x)) = 0

Step 3: Group like terms Combine the terms with sin^2(x) and cos^2(x):

3sin^2(x) + 5cos^2(x) - 2cos^2(x) + 2sin^2(x) + 8sin(x)cos(x) = 0

Step 4: Simplify further Combine the terms with sin^2(x) and cos^2(x) again:

5sin^2(x) + 3cos^2(x) + 8sin(x)cos(x) = 0

Step 5: Use the Pythagorean identity Since sin^2(x) + cos^2(x) = 1, we can express sin^2(x) and cos^2(x) in terms of the other:

5(1 - cos^2(x)) + 3cos^2(x) + 8sin(x)cos(x) = 0

Step 6: Simplify the equation Distribute the 5 and rearrange:

5 - 5cos^2(x) + 3cos^2(x) + 8sin(x)cos(x) = 0

Combine the terms:

5 + 3cos^2(x) + 8sin(x)cos(x) = 0

Step 7: Solve for cos(x) To solve for cos(x), we can treat the equation as a quadratic in cos(x):

3cos^2(x) + 8sin(x)cos(x) + 5 = 0

This is a quadratic equation in cos(x). Let's temporarily replace sin(x) with a variable, say, "a", to solve the quadratic equation:

3cos^2(x) + 8a*cos(x) + 5 = 0

Now, solve the quadratic equation for cos(x). The solutions for cos(x) will be expressed in terms of "a".

Step 8: Substitute back sin(x) Now that we have the solutions for cos(x) in terms of "a", substitute back sin(x) for "a" to get the final solutions for x.

Keep in mind that trigonometric equations often have multiple solutions due to their periodic nature, so the final answer may have more than one solution depending on the range of x you are interested in.