Question

2cos^2x-7cosx=2sin^2x

Answer 1

$2\cos^2x-7\cos x=2\sin^2x\\ 2(\cos^2x-\sin^2x)-7\cos x=0\\ 2(2\cos^2x-1)-7\cos x=0\\ 4\cos^2x-7\cos x-2=0$

Пусть cos x = t, причем |t|≤1, тогда получаем
$4t^2-7t-2=0\\ D=b^2-4ac=49+16\cdot 2=81 \\ t_1=-0.25$
$t_2=2$ - не удовлетворяет условию

Возвращаемся к замене
$\cos x=-0.25\\ x=\pm\arccos(-0.25)+2 \pi n,n \in Z$

Answer 2

To solve the equation 2cos^2x - 7cosx = 2sin^2x, we can use some trigonometric identities to simplify it.

Let's start by using the identity sin^2x + cos^2x = 1. Rearranging this identity, we have sin^2x = 1 - cos^2x.

Substituting this into the original equation, we get:

2cos^2x - 7cosx = 2(1 - cos^2x).

Expanding the right side of the equation, we have:

2cos^2x - 7cosx = 2 - 2cos^2x.

Combining like terms, we obtain:

4cos^2x - 7cosx - 2 = 0.

Now, we have a quadratic equation in terms of cosx. Let's solve it by factoring or using the quadratic formula.

Factoring the quadratic equation, we have:

(2cosx + 1)(2cosx - 2) = 0.

Setting each factor to zero and solving for cosx, we get:

2cosx + 1 = 0 ---> cosx = -1/2,

2cosx - 2 = 0 ---> cosx = 1.

Now, let's find the corresponding values of sinx for each solution of cosx.

For cosx = -1/2, we can use the identity sin^2x + cos^2x = 1 to find sinx:

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

Taking the square root of both sides, we have:

sinx = ±√(3/4) = ±√3/2.

So, when cosx = -1/2, sinx can be either -√3/2 or √3/2.

For cosx = 1, we can again use the identity sin^2x + cos^2x = 1:

sin^2x + 1^2 = 1, sin^2x + 1 = 1, sin^2x = 0.

Taking the square root of both sides, we have:

sinx = ±0 = 0.

So, when cosx = 1, sinx is equal to 0.

Therefore, the solutions to the equation 2cos^2x - 7cosx = 2sin^2x are:

1. cosx = -1/2, sinx = -√3/2,
2. cosx = -1/2, sinx = √3/2,
3. cosx = 1, sinx = 0.

These are the values of x that satisfy the given equation.