7cosx+5sinx=2k+1

6
35
3
14

To solve the equation 7cos(x) + 5sin(x) = 2k + 1, where k is a constant, we need to find the possible values of x.

First, let's rearrange the equation:

7cos(x) + 5sin(x) = 2k + 1

We can rewrite the left side of the equation using the trigonometric identity:

a*cos(x) + b*sin(x) = sqrt(a^2 + b^2)*cos(x - theta)

Where theta is the angle whose tangent is b/a. In this case, a = 7 and b = 5. So we have:

7cos(x) + 5sin(x) = sqrt(7^2 + 5^2)*cos(x - theta)

7cos(x) + 5sin(x) = sqrt(74)*cos(x - theta)

Now, we can rewrite the equation as:

sqrt(74)*cos(x - theta) = 2k + 1

Next, we divide both sides of the equation by sqrt(74):

cos(x - theta) = (2k + 1) / sqrt(74)

To find the possible values of x, we need to find the values of (x - theta) that satisfy the equation. We can use the inverse cosine function to find these values.

x - theta = arccos((2k + 1) / sqrt(74))

x = arccos((2k + 1) / sqrt(74)) + theta

Now, we can substitute the given values of k to find the corresponding values of x.

For k = 6:
x = arccos((2(6) + 1) / sqrt(74)) + theta

For k = 35:
x = arccos((2(35) + 1) / sqrt(74)) + theta

For k = 3:
x = arccos((2(3) + 1) / sqrt(74)) + theta

For k = 14:
x = arccos((2(14) + 1) / sqrt(74)) + theta

To evaluate these expressions, you can use a scientific calculator or an online trigonometry calculator. Simply input the corresponding values of k and evaluate the expression.