To solve the equation 3cos^2(x) = 7(sin(x) + 1), we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to substitute for cos^2(x) in terms of sin(x).
3(1 - sin^2(x)) = 7(sin(x) + 13 - 3sin^2(x) = 7sin(x) + 7
Rearranging the terms, we get3sin^2(x) + 7sin(x) - 4 = 0
Now, let's solve this quadratic equation for sin(x). We can factor it, or use the quadratic formula:
sin(x) = [-7 ± √(7^2 - 4(3)(-4))] / 2(3sin(x) = [-7 ± √(49 + 48)] / sin(x) = [-7 ± √97] / 6
Therefore, the solutions for sin(x) aresin(x) = (-7 + √97) / sin(x) = (-7 - √97) / 6
Now that we have the values of sin(x), we can find the corresponding values of cos(x) using the Pythagorean identity:
cos(x) = ±√(1 - sin^2(x))
For the positive value of sin(x)cos(x) = ±√(1 - (-7 + √97)^2 / 3For the negative value of sin(x)cos(x) = ±√(1 - (-7 - √97)^2 / 36
These are the solutions to the equation 3cos^2(x) = 7(sin(x) + 1).
To solve the equation 3cos^2(x) = 7(sin(x) + 1), we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to substitute for cos^2(x) in terms of sin(x).
3(1 - sin^2(x)) = 7(sin(x) + 1
3 - 3sin^2(x) = 7sin(x) + 7
Rearranging the terms, we get
3sin^2(x) + 7sin(x) - 4 = 0
Now, let's solve this quadratic equation for sin(x). We can factor it, or use the quadratic formula:
sin(x) = [-7 ± √(7^2 - 4(3)(-4))] / 2(3
sin(x) = [-7 ± √(49 + 48)] /
sin(x) = [-7 ± √97] / 6
Therefore, the solutions for sin(x) are
sin(x) = (-7 + √97) /
sin(x) = (-7 - √97) / 6
Now that we have the values of sin(x), we can find the corresponding values of cos(x) using the Pythagorean identity:
cos(x) = ±√(1 - sin^2(x))
For the positive value of sin(x)
cos(x) = ±√(1 - (-7 + √97)^2 / 3
For the negative value of sin(x)
cos(x) = ±√(1 - (-7 - √97)^2 / 36
These are the solutions to the equation 3cos^2(x) = 7(sin(x) + 1).