Question

2sin^2 ×X-cosX-1=0 [3π;4π]

Answer 1

2(1-сos²x)-cosx-1=0
2-2cos²x-cosx-1=0
cosx=a
2a²+a-1=0
D=1+8=9
a1=(---3)/4=-1⇒cosx=-1⇒x=π+2πn,n∈z
3π≤π+2πn≤4π
3≤1+2n≤4
2≤2n≤3
1≤n≤1,5
n=1⇒x=π+2π=3π
a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πn
3π≤-π/3+2πn≤4π
9≤-1+6n≤12
10≤6n≤13
10/6≤n≤13/6
n=2⇒x=-π/3+4π=11π/3
3π≤π/3+2πn≤4π
9≤1+6n≤12
8≤6n≤11
8/6≤n≤11/6
нет решения

Answer 2

To solve this equation, we can use a substitution. Let's substitute u = sin(x). Then, we have:

2u^2 - cos(x) - 1 = 0

We can use the identity cos(x) = √(1 - sin^2(x)) to express cos(x) in terms of u:

2u^2 - √(1 - u^2) - 1 = 0

Let's multiply both sides by the conjugate of the denominator to get rid of the square root:

(2u^2 - 1) + u^2√(1 - u^2) = 0

Now, we can isolate the square root on one side:

u^2√(1 - u^2) = 1 - 2u^2

Let's square both sides:

u^4(1 - u^2) = (1 - 2u^2)^2

Expanding the right-hand side, we get:

u^4 - 2u^6 = 1 - 4u^2 + 4u^4

Moving all the terms to one side, we obtain:

2u^6 - 5u^4 + 4u^2 - 1 = 0

This is a sixth-degree polynomial equation in u. We can try to solve it using numerical methods or approximations. However, in this case, we can use the fact that the solutions are restricted to the interval [sin(3π), sin(4π)] = [0, 0], since sin(x) is non-negative in that interval. Therefore, u = 0 is a solution of the equation.

To find the other solutions, we can divide the polynomial by u^2 - 1, which is a factor of it, to obtain a quartic equation:

2u^4 + 3u^3 - u^2 - 3u + 1 = 0

We can factor this equation as follows:

(2u^2 - u - 1)(u^2 + 2u - 1) = 0

The solutions of the quartic equation are u = (1 ± √3)/4 and u = -1 ± √2. However, only u = (1 + √3)/4 and u = -1 + √2 are in the interval [0, 1], which corresponds to the range of sin(x) in [3π, 4π]. Therefore, the solutions of the original equation in the interval [3π, 4π] are:

sin(x) = 0, (1 + √3)/4, and arcsin(-1 + √2).