Question

Sinxcos2x+cosxsin2x>1/2​

Answer 1

Ответ:

$\sin(x + 2x) = \frac{1}{2}$

$\sin(3x) = \frac{1}{2}$

$3x = ( - 1)^{n} \frac{\pi}{6} + \pi \times n$

$x = { (- 1)}^{n} \frac{\pi}{18} + \frac{\pi \times n}{3}$

Answer 2

To solve the inequality sin(x)cos(2x) + cos(x)sin(2x) > 1/2, we can simplify the left-hand side and analyze the resulting expression.

Using the angle addition formula for sine and cosine, we have:

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

Now, let's simplify further using the identity cos^2(x) - sin^2(x) = cos(2x):

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

Now, the inequality becomes:

sin(x) + cos(x)cos(2x) > 1/2

To solve this inequality, we need to consider different cases and analyze them separately.

Case 1: 0 < x < π/2 (i.e., x is in the first quadrant)

In this case, both sin(x) and cos(x) are positive. We can rewrite the inequality as:

sin(x) + cos(x)cos(2x) > 1/2

Since both sin(x) and cos(x) are positive, we have sin(x) > 0 and cos(x) > 0. Therefore, the inequality reduces to:

sin(x) + cos(x)cos(2x) > 1/2

Case 2: π/2 < x < π (i.e., x is in the second quadrant)

In this case, sin(x) is positive, but cos(x) is negative. We can rewrite the inequality as:

sin(x) + cos(x)cos(2x) > 1/2

Since sin(x) > 0 and cos(x) < 0, the inequality becomes:

sin(x) + cos(x)cos(2x) > 1/2

Case 3: π < x < 3π/2 (i.e., x is in the third quadrant)

In this case, both sin(x) and cos(x) are negative. We can rewrite the inequality as:

sin(x) + cos(x)cos(2x) > 1/2

Since both sin(x) and cos(x) are negative, we have sin(x) < 0 and cos(x) < 0. Therefore, the inequality reduces to:

sin(x) + cos(x)cos(2x) > 1/2

Case 4: 3π/2 < x < 2π (i.e., x is in the fourth quadrant)

In this case, sin(x) is negative, but cos(x) is positive. We can rewrite the inequality as:

sin(x) + cos(x)cos(2x) > 1/2

Since sin(x) < 0 and cos(x) > 0, the inequality becomes:

sin(x) + cos(x)cos(2x) > 1/2

By analyzing the inequality in each case, we can see that the solution will depend on the specific values of x within each interval. Further analysis or numerical methods may be required to determine the exact range of x for which the inequality holds true.