Question

Sin2x + 1 = cosx + 2sinx

Answer 1

Ответ:

$2\pi k, ~k\in\mathbb {Z} ; (-1)^{n} \cdot\dfrac{\pi }{6}+\pi n, ~n\in\mathbb {Z}$

Объяснение:

$sin2x+1=cosx+2sinx.$

Воспользуемся формулой синуса двойного угла.

$sin2x=2sinxcosx.\\2sinxcosx+1=cosx+2sinx;\\2sinxcosx+1-cosx-2sinx=0;\\2sinx(cosx-1)-(cosx-1)=0;\\(cosx-1)(2sinx-1)=0;\\ \left [\begin{array}{l}cosx-1 = 0, \\ 2sinx-1= 0; \end{array} \right.\Leftrightarrow \left [\begin{array}{l} cosx=1 , \\ sinx = \dfrac{1}{2} ; \end{array} \Leftrightarrow \left [\begin{array}{l} x = 2\pi k, ~k\in\mathbb {Z} \\ x = (-1)^{n} \cdot\dfrac{\pi }{6}+\pi n, ~n\in\mathbb {Z} \end{array} \right.\right.$

Answer 2

To solve the trigonometric equation sin(2x) + 1 = cos(x) + 2sin(x), you can use trigonometric identities to simplify and rearrange the equation. Here's how to do it step by step:

1. Start with the equation: sin(2x) + 1 = cos(x) + 2sin(x)

2. Use the double angle identity for sine to express sin(2x) in terms of sin(x) and cos(x): sin(2x) = 2sin(x)cos(x)

3. Substitute the expression for sin(2x) into the equation: 2sin(x)cos(x) + 1 = cos(x) + 2sin(x)

4. Move all terms involving sin(x) to one side of the equation and those involving cos(x) to the other side: 2sin(x)cos(x) - 2sin(x) - cos(x) + 1 = 0

5. Factor out common terms: 2sin(x)(cos(x) - 1) - (cos(x) - 1) = 0

6. Notice that (cos(x) - 1) appears in both terms. Factor it out: (cos(x) - 1)(2sin(x) - 1) = 0

Now you have two factors, and you can set each factor equal to zero to solve for x:

1. cos(x) - 1 = 0 cos(x) = 1

2. 2sin(x) - 1 = 0 2sin(x) = 1 sin(x) = 1/2

Solving for x in the first equation, you get: x = 0

Solving for x in the second equation, you get: x = π/6 + 2πn, where n is an integer.

So, the solutions to the equation sin(2x) + 1 = cos(x) + 2sin(x) are: x = 0 x = π/6 + 2πn, where n is an integer.