Question

Sinx ×cosx=-√3/2sinx​

Answer 1

Ответ:

Пошаговое объяснение:

sinxcosx+V3/2sinx=0,  sinx(cosx+V3/2)=0,

sinx=0,  x=pn,    cosx+V3/2=0,  cosx=-V3/2,  x=+-5p/6+2pn,

n   E   Z

Answer 2

To solve the equation, we need to find the values of x that satisfy the given equation: sin(x) × cos(x) = -√3/2 × sin(x).

Let's first rewrite the equation:

sin(x) × cos(x) + √3/2 × sin(x) = 0

Next, factor out sin(x) from the left-hand side:

sin(x) (cos(x) + √3/2) = 0

Now, we have two possibilities for the equation to be true:

1. sin(x) = 0
2. cos(x) + √3/2 = 0

For the first case, sin(x) = 0, which occurs when x is any multiple of π, i.e., x = 0, π, 2π, 3π, ...

For the second case, we need to solve for cos(x):

cos(x) = -√3/2

To find the solutions for cos(x) = -√3/2, we can look at the unit circle for reference. At which angles does the cosine function have a value of -√3/2? The unit circle shows that cos(π/6) = cos(30°) = √3/2, and since cos(x) is an even function, we also have cos(11π/6) = cos(330°) = √3/2.

However, we want the solutions for cos(x) = -√3/2, which means we need to consider the angles in the second and third quadrants where cos(x) is negative. In the second quadrant, cos(π - θ) = -cos(θ), and in the third quadrant, cos(θ + π) = -cos(θ). So, the solutions in the second and third quadrants are π - π/6 = 5π/6 and π + π/6 = 7π/6, respectively.

Therefore, the solutions for the second case, cos(x) + √3/2 = 0, are x = 5π/6 and x = 7π/6.

In summary, the solutions to the equation sin(x) × cos(x) = -√3/2 sin(x) are:

x = 0, π, 2π, 3π, ... (any multiple of π)

and

x = 5π/6, 7π/6 (corresponding to 150° and 210° in degrees, respectively).