Question

6sin^2 x+(3/2)sin2x-2=5cos^2 x

Answer 1

$6sin^2x+ \dfrac{3}{2} sin2x-2=5cos^2x|*2 \\ 12sin^2x-3sin2x-4=10cos^2x \\ 12sin^2x-3*2sinx*cosx-4(sin^2x+cos^2x)=10cos^2x \\ 12sin^2x-6sinx*cosx-4sin^2x-4cos^2x-10cos^2x=0 \\ 8sin^2x-6sinx*cosx-14cos^2x=0|:2 \\ 4sin^2x-3sin*cosx-7cos^2x=0|:cos^2x \\ 4tg^2x-3tgx-7=0$

Пусть tg x = t (t ∈ R), тогда имеем

$4t^2-3t-7=0 \\ D=b^2-4ac=(-3)^2-4*4*(-7)=121 \\ t_1=-1; \\ t_2=1.75$

Обратная замена

$tgx=-1 \\ x_1=-arctg1+ \pi n, n \in Z \\ x_1=- \dfrac{ \pi }{4} + \pi n, n \in Z \\ \\ tgx=1.75 \\ x_2=arctg1.75+ \pi n,n \in Z$

Answer 2

To solve the equation 6sin^2(x) + (3/2)sin(2x) - 2 = 5cos^2(x), we can use trigonometric identities to simplify the equation and then solve for the values of x.

Let's start by using the double-angle identity for sine, which states that sin(2x) = 2sin(x)cos(x). We can substitute this into the equation:

6sin^2(x) + (3/2)(2sin(x)cos(x)) - 2 = 5cos^2(x)

Now, let's simplify further:

6sin^2(x) + 3sin(x)cos(x) - 2 = 5cos^2(x)

Next, let's use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace sin^2(x) and cos^2(x):

6(1 - cos^2(x)) + 3sin(x)cos(x) - 2 = 5(1 - sin^2(x))

Distribute and rearrange the equation:

6 - 6cos^2(x) + 3sin(x)cos(x) - 2 = 5 - 5sin^2(x)

Combine like terms:

-6cos^2(x) + 3sin(x)cos(x) - 6sin^2(x) = -4

Divide the entire equation by -1 to make the coefficient of the quadratic term positive:

6cos^2(x) - 3sin(x)cos(x) + 6sin^2(x) = 4

Now, we can simplify the equation further:

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

Since sin^2(x) = 1 - cos^2(x), we can substitute this into the equation:

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

Distribute and rearrange:

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

-sin(x)cos(x) + 2 = 4/3

Now, we have a simplified equation:

-sin(x)cos(x) = 4/3 - 2

-sin(x)cos(x) = 4/3 - 6/3

-sin(x)cos(x) = -2/3

At this point, it becomes more challenging to find exact solutions algebraically. However, we can approximate solutions using numerical methods or calculators.