Question

5sin²x+3sin2x-3cos²2x=4

Answer 1

$5sin^2x+3sin2x-3cos^2x=4\\\\5sin^2x+3\cdot 2\, sinx\, cosx-3cos^2x=4(sin^2x+cos^2x)\\\\sin^2x+6\, sinx\, cosx-7cos^2x=0\; |:cos^2x\ne 0\\\\tg^2x+6\, tgx-7=0\\\\D=64\; ,\; \; (tgx)_1=-7\; ,\; \; (tgx)_2=1\\\\\underline {x_1=-arctg\, 7+\pi n\; ,\; n\in Z}\\\\\underline {x_2= \frac{\pi }{4}+\pi k\; ,\; k\in Z}$

Answer 2

We can start by using the identity cos²θ + sin²θ = 1 to replace cos²2x with 1 - sin²2x:

5sin²x + 3sin2x - 3(1 - sin²2x) = 4

Simplifying and rearranging terms, we get:

5sin²x + 3sin2x - 3 + 3sin²2x = 4

Combining like terms:

8sin²x + 3sin2x - 3 = 4 - 3sin²2x

Using the identity sin2θ = 2sinθcosθ, we can replace sin2x with 2sinxcosx:

8sin²x + 6sinxcosx - 3 = 4 - 3(2sinxcosx)²

Simplifying:

8sin²x + 6sinxcosx - 3 = 4 - 12sin²x cos²x

Moving all the terms to one side:

12sin²x cos²x + 6sinxcosx - 11 = 0

We can recognize this as a quadratic equation in terms of sinx cosx, so we can use the quadratic formula:

sinx cosx = [-6 ± sqrt((6²) - 4(12)(-11))]/(2*12)

sinx cosx = [-6 ± sqrt(636)]/24

sinx cosx ≈ -0.553 or sinx cosx ≈ 0.34

Now we have two cases to consider:

Case 1: sinx cosx ≈ -0.553

We know that sinx and cosx have opposite signs, so we can write:

sinx ≈ -sqrt(0.553) ≈ -0.743 and cosx ≈ sqrt(0.553) ≈ 0.744

Using the identity sin²θ + cos²θ = 1, we can check that this solution satisfies the original equation.

Case 2: sinx cosx ≈ 0.34

We know that sinx and cosx have the same sign, so we can write:

sinx ≈ sqrt(0.34) ≈ 0.583 and cosx ≈ sqrt(0.34) ≈ 0.583

Using the identity sin²θ + cos²θ = 1, we can check that this solution does not satisfy the original equation.

Therefore, the only solution to the equation is:

sinx ≈ -0.743 and cosx ≈ 0.744

We can use inverse trigonometric functions to find the value of x:

x ≈ -0.862 radians + 2πn or x ≈ 2.004 radians + 2πn, where n is an integer.