Question

Sin^2x cosx+2sin^2x-0,5cosx-1=0

Answer 1

Ответ:

$x_{1}=(-1)^{n+1}*\frac{\pi}{4}+\pi n,$n∈Z

$x_{2}=(-1)^{n}*\frac{\pi}{4}+\pi n,$n∈Z

Пошаговое объяснение:

sin²x*cosx+2sin²x-0,5cosx-1=0

(sin²x*cosx-0,5cosx)+(2sin²x-1)=0

cosx*(sin²x-0,5)+2*(sin²x-0,5)=0

(sin²x-0,5)*(cosx+2)=0

sin²x-0,5=0  или cosx+2=0

sin²x=0,5  или cosx= - 2, -2∉[-1;1], => решений нет

$sin^{2} x=\frac{1}{2}$

$sinx=-+\sqrt{\frac{1}{2}}$

$\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

1. $sinx=-\frac{\sqrt{2}}{2}$

$x=(-1)^{n}*arcsin(-\frac{\sqrt{2} }{2})+\pi n,$n∈Z

$x=(-1)^{n+1}*\frac{\pi}{4}+\pi n,$n∈Z

2. $sinx=\frac{\sqrt{2}}{2}$

$x=(-1)^{n}*\frac{\pi}{4}+\pi n,$n∈X

Answer 2

Let's solve the given equation for x:

sin^2(x) cos(x) + 2sin^2(x) - 0.5cos(x) - 1 = 0

To make this equation more manageable, let's use a substitution: Let y = sin(x).

Now, we can rewrite the equation as follows:

y^2 * cos(x) + 2y^2 - 0.5 * cos(x) - 1 = 0

Now, let's try to factor this equation. Group the terms that have cos(x) and the terms that do not involve cos(x):

(cos(x) + 2)y^2 - (0.5 * cos(x) + 1) = 0

Factor out the common terms:

(cos(x) + 2)(y^2 - 0.5) = 0

Now, we have two possible cases:

1. cos(x) + 2 = 0
2. y^2 - 0.5 = 0

Let's solve each case:

Case 1: cos(x) + 2 = 0

Subtract 2 from both sides:

cos(x) = -2

The range of cosine function is -1 to 1, and it does not equal -2. Therefore, there is no solution for this case.

Case 2: y^2 - 0.5 = 0

Add 0.5 to both sides:

y^2 = 0.5

Take the square root of both sides:

y = ±√(0.5)

Now, remember that y = sin(x), so we have:

sin(x) = ±√(0.5)

To find the values of x, we need to find the arcsin of ±√(0.5). Using a calculator, we get:

x = arcsin(±√(0.5)) ≈ ±45° + k*180°, where k is an integer.

So, the solutions for x are approximately:

x ≈ 45° + k180° and x ≈ -45° + k180°, where k is an integer.