Question

Sin2x+cosx=2sinx+ Подробно если можно , зарание спасибо.

Answer 1

Sin2x+√(3)cosx-2sinx-√(3)=0
2sinxcosx-2sinx+√(3)cosx-√(3)=0
2sinx(cosx-1)+√(3)(cosx-1)=0
(cosx-1)(2sinx+√(3))=0
cosx=1
x=2pi*n
sinx=-√(3)/2
x=((-1)^n)*4pi/3+2pi*k
k∈Z, n∈Z

Answer 2

Solving the Equation sin(2x) + cos(x) = 2sin(x)

To solve the equation sin(2x) + cos(x) = 2sin(x), we can use trigonometric identities and algebraic manipulation.

First, let's express sin(2x) in terms of sin(x) and cos(x) using the double-angle identity for sine: sin(2x) = 2sin(x)cos(x)

Now, we can substitute sin(2x) with 2sin(x)cos(x) in the original equation: 2sin(x)cos(x) + cos(x) = 2sin(x)

Next, we can rearrange the terms to solve for sin(x) and cos(x). Let's subtract 2sin(x) from both sides: 2sin(x)cos(x) - 2sin(x) = -cos(x)

Now, we can factor out 2sin(x) from the left-hand side: 2sin(x)(cos(x) - 1) = -cos(x)

Finally, we can solve for sin(x) by dividing both sides by (cos(x) - 1): sin(x) = -cos(x) / (2(cos(x) - 1))

This provides an expression for sin(x) in terms of cos(x). Further simplification may be possible, but this is the general approach to solving the given equation.

Let me know if you need further assistance with this or any other topic!