Next, we can rewrite cos^2(3x) in terms of sin(3x) using the identity cos^2(x) + sin^2(x) = 1:
1 - sin^2(3x) + 2sin(3x) = 3
Now, let's substitute y = sin(3x) to simplify the equation:
1 - y^2 + 2y = 3
Rearranging terms gives us:
y^2 - 2y - 2 = 0
Now we need to solve this quadratic equation for y. We can use the quadratic formula:
y = [ 2 ± sqrt(2^2 - 41(-2)) ] / 2* y = [ 2 ± sqrt(4 + 8) ] / y = [ 2 ± sqrt(12) ] / y = [2 ± 2√3] / y = 1 ± √3
Therefore, sin(3x) = 1 ± √3
To find the values of x, we need to determine the possible values of sin(3x) given that it equals 1 ± √3. Since the range of sin function is -1 to 1, we can see that sin(3x) can only be 1 + √3.
So, sin(3x) = 1 + √3
From here, we can solve for x by taking the inverse sine of both sides:
To solve the equation cos^2(3x) + 3sin(3x) = 3, we can use the trigonometric identity cos^2(x) + sin^2(x) = 1.
First, we rewrite the equation using this identity:
cos^2(3x) + 3sin(3x) = cos^2(3x) + sin(3x) + sin(3x) = cos^2(3x) + 2sin(3x) = 3
Next, we can rewrite cos^2(3x) in terms of sin(3x) using the identity cos^2(x) + sin^2(x) = 1:
1 - sin^2(3x) + 2sin(3x) = 3
Now, let's substitute y = sin(3x) to simplify the equation:
1 - y^2 + 2y = 3
Rearranging terms gives us:
y^2 - 2y - 2 = 0
Now we need to solve this quadratic equation for y. We can use the quadratic formula:
y = [ 2 ± sqrt(2^2 - 41(-2)) ] / 2*
y = [ 2 ± sqrt(4 + 8) ] /
y = [ 2 ± sqrt(12) ] /
y = [2 ± 2√3] /
y = 1 ± √3
Therefore, sin(3x) = 1 ± √3
To find the values of x, we need to determine the possible values of sin(3x) given that it equals 1 ± √3. Since the range of sin function is -1 to 1, we can see that sin(3x) can only be 1 + √3.
So, sin(3x) = 1 + √3
From here, we can solve for x by taking the inverse sine of both sides:
3x = sin^(-1)(1 + √3)
x = sin^(-1)(1 + √3) / 3
Therefore, x = sin^(-1)(1 + √3) / 3.