Question

4sin(3x)sinx-2cos(2x)+1=0

Answer 1

$4*0.5*(cos2x-cos4x)-2cos2x+1=0$
$2cos2x-2cos4x-2cos2x+1=0$
$-2cos4x+1=0$
$cos4x=0.5$
$4x=+-\frac{ \pi }{3} +2 \pi k$
$x=+-\frac{ \pi }{12} +\frac{ \pi k}{2}$, k∈Z

Воспользовалась формулой:
$sina*sinb=0.5*(cos(a-b)-cos(a+b))$

Answer 2

We can start by applying some trigonometric identities to simplify the equation.

First, we can use the double angle identity for cosine:

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

We can also use the product-to-sum identity for sine:

2sin(A)sin(B) = cos(A-B) - cos(A+B)

Using these identities, we can rewrite the equation as:

4sin(3x)sinx - 2(2cos^2(x) - 1) + 1 = 0

Simplifying further:

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

Now, we can use the double angle identity for sine:

sin(2x) = 2sin(x)cos(x)

sin(3x) = 3sin(x) - 4sin^3(x)

Substituting these identities into the equation:

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

Simplifying:

12sin^2(x) - 16sin^4(x) - 4cos^2(x) + 3 = 0

Using the Pythagorean identity:

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

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

Substituting into the equation:

12sin^2(x) - 16sin^4(x) - 4(1 - sin^2(x)) + 3 = 0

Expanding and simplifying:

-16sin^4(x) + 12sin^2(x) - 1 = 0

Now, we can substitute y = sin^2(x) and rewrite the equation in terms of y:

-16y^2 + 12y - 1 = 0

We can solve for y using the quadratic formula:

y = [ -b ± sqrt(b^2 - 4ac) ] / 2a

where a = -16, b = 12, and c = -1.

Plugging in the values:

y = [ -12 ± sqrt(12^2 - 4(-16)(-1)) ] / 2(-16) y = [ -3 ± sqrt(5) ] / 4

Since y = sin^2(x), we need to take the square root of y to find sin(x):

sin(x) = ± sqrt([ -3 ± sqrt(5) ] / 4)

Therefore, the solutions are:

sin(x) = sqrt([ -3 + sqrt(5) ] / 4), sin(x) = -sqrt([ -3 + sqrt(5) ] / 4), sin(x) = sqrt([ -3 - sqrt(5) ] / 4), and sin(x) = -sqrt([ -3 - sqrt(5) ] / 4)

Note that some of these solutions may not be valid, depending on the range of values that x is supposed to take.