Question

1) sin2x-0,5=0 2)6-5sinx=1 3)4-3sinx=7 4)-9cosx+4=4

Answer 1

Решение на фотографии

Answer 2

To solve the equations, we'll find the values of 'x' that satisfy each equation. Let's solve them one by one:

1. sin(2x) - 0.5 = 0:

To solve this equation, first, let's isolate sin(2x) by adding 0.5 to both sides:

sin(2x) = 0.5

Now, to find the values of 'x', take the inverse sine (sin^(-1)) of both sides:

2x = sin^(-1)(0.5)

Now, solve for 'x':

x = (sin^(-1)(0.5)) / 2

Using a calculator, we find the value of 'x':

x ≈ 0.5236 radians (approximately 30 degrees)

2. 6 - 5sin(x) = 1:

Let's isolate sin(x) by moving 6 to the right side:

-5sin(x) = 1 - 6

-5sin(x) = -5

Now, divide both sides by -5:

sin(x) = 1

To find the values of 'x', take the inverse sine (sin^(-1)) of both sides:

x = sin^(-1)(1)

Using a calculator, we find the value of 'x':

x = π/2 (pi/2) or 90 degrees

3. 4 - 3sin(x) = 7:

Let's isolate sin(x) by moving 4 to the right side:

-3sin(x) = 7 - 4

-3sin(x) = 3

Now, divide both sides by -3:

sin(x) = -1

To find the values of 'x', take the inverse sine (sin^(-1)) of both sides:

x = sin^(-1)(-1)

Using a calculator, we find the value of 'x':

x = -π/2 (-pi/2) or -90 degrees

4. -9cos(x) + 4 = 4:

Let's isolate cos(x) by moving 4 to the right side and then divide by -9:

-9cos(x) = 4 - 4

-9cos(x) = 0

cos(x) = 0

To find the values of 'x', take the inverse cosine (cos^(-1)) of both sides:

x = cos^(-1)(0)

Using a calculator, we find the value of 'x':

x = π/2 (pi/2) or 90 degrees

Note: Remember that trigonometric functions have periodicity, so there are infinitely many solutions to these equations. The solutions provided above are principal solutions within a specific range. If you need a broader range of solutions, you can add or subtract multiples of the period (2π) to the angles provided above.