Question

(1+2sinx)sinx=sin2x+cosx

Answer 1

$(1 + 2sinx)sinx = sin2x + cosx \\ (1 + 2sinx)sinx = 2sinxcosx + cosx \\ (1 + 2sinx)sinx = (1 + 2sinx)cosx \\ (1 + 2sinx)(sinx - cosx) = 0 \\ \\ 1) \: 1 + 2sinx = 0 \\ sinx = - \frac{1}{2} \\ x = - \frac{\pi}{6} + 2\pi \: n \\ x = - \frac{5\pi}{6} + 2\pi \: k$
n и k принадлежат Z


$2) \: \: sinx - cosx = 0$
Обе части разделим на cosx =/ 0

$tgx - 1 = 0 \\ tgx = 1 \\ x = \frac{\pi}{4} + \pi \: m$
m принадлежит Z



ОТВЕТ: - п/6 + 2пn ; - 5п/6 + 2пk ; п/4 + пm , n , k и m принадлежат Z

Answer 2

task/30079176    ( 1 + 2sinx)sinx = sin2x+cosx .

решение (1+2sinx)sinx=sin2x+cosx ⇔ (1+2sinx)sinx=2sinxcosx +cosx ⇔(1+2sinx)sinx =(2sinx + 1 )cosx ⇔ (1+2sinx)sinx -(2sinx + 1 )cosx = 0 ⇔ (2sinx+1)(sinx - cosx) =0⇔ [2sinx +1 =0 ; sinx - cosx =0. ⇔

[ sinx  = -1/2 ; sinx =cosx. ⇔ [ sinx  = -1/2 ; tgx =1 . ⇔

[ x = (-1)ⁿ⁻¹(π/6) +πn , x =π/4 +πn , n∈ℤ .

Answer 3

To solve the trigonometric equation (1 + 2sin(x))sin(x) = sin(2x) + cos(x), you can follow these steps:

1. Expand the trigonometric expressions on both sides of the equation using trigonometric identities:

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

   sin(2x) = 2sin(x)cos(x)

   cos(x) is already in its standard form.

2. Substitute the expanded expressions back into the original equation:

   sin(x) + 2 - 2cos^2(x) = 2sin(x)cos(x) + cos(x)

3. Move all terms to one side of the equation to set it equal to zero:

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

4. Group like terms:

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

5. Factor out common terms:

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

6. Notice that (1 - 2cos(x)) is a common factor, so you can factor it out:

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

7. Now, you have a quadratic equation in the form of (1 - 2cos(x)):

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

8. Set each factor equal to zero:

   1 - 2cos(x) = 0 => 2cos(x) = 1 => cos(x) = 1/2 sin(x) + cos(x) + 2 = 0

9. Solve for cos(x) = 1/2:

   Since cos(60°) = 1/2 and cos(300°) = 1/2, we have:

   x = 60° + 360°n x = 300° + 360°n

10. Now, solve the remaining equation for sin(x) + cos(x) + 2 = 0:

sin(x) + cos(x) + 2 = 0

Subtract cos(x) from both sides:

sin(x) + 2 = -cos(x)

Square both sides to get rid of the square root:

sin^2(x) + 4sin(x) + 4 = cos^2(x)

Use the trigonometric identity sin^2(x) + cos^2(x) = 1:

4sin(x) + 4 = 1 - 4sin(x)

Add 4sin(x) to both sides:

8sin(x) + 4 = 1

Subtract 4 from both sides:

8sin(x) = -3

Divide by 8:

sin(x) = -3/8

To find the solutions, use the inverse sine function:

x = arcsin(-3/8)

However, you may want to use a calculator to find the approximate values of x since arcsin(-3/8) is not a standard angle, and it will give you a decimal value.