To solve this trigonometric equation, we can first rewrite it using the double angle formula for sine, which states that sin(2x) = 2sin(x)cos(x).
So, we have:
2sin²x + 1.5(2sin(x)cos(x)) - 3cos²x = 12sin²x + 3sin(x)cos(x) - 3cos²x = 1
Now, we can use the Pythagorean identity sin²x + cos²x = 1 to substitute for one of the trigonometric functions. Let's substitute for sin²x:
2(1-cos²x) + 3sin(x)cos(x) - 3cos²x = 12 - 2cos²x + 3sin(x)cos(x) - 3cos²x = 1-5cos²x + 3sin(x)cos(x) - 1 = 0
Now, we can rewrite the equation in terms of only cos(x):
5cos²x - 3sin(x)cos(x) + 1 = 0
Solving this quadratic equation in cos(x) might be a bit tricky. We might need to use the quadratic formula:
cos(x) = [3sin(x) ± sqrt((3sin(x))² - 451)] / 10
We can then use the Pythagorean identity sin²x + cos²x = 1 to find the corresponding sine values for the solutions of cos(x).
I hope this helps! Let me know if you have any questions.
To solve this trigonometric equation, we can first rewrite it using the double angle formula for sine, which states that sin(2x) = 2sin(x)cos(x).
So, we have:
2sin²x + 1.5(2sin(x)cos(x)) - 3cos²x = 1
2sin²x + 3sin(x)cos(x) - 3cos²x = 1
Now, we can use the Pythagorean identity sin²x + cos²x = 1 to substitute for one of the trigonometric functions. Let's substitute for sin²x:
2(1-cos²x) + 3sin(x)cos(x) - 3cos²x = 1
2 - 2cos²x + 3sin(x)cos(x) - 3cos²x = 1
-5cos²x + 3sin(x)cos(x) - 1 = 0
Now, we can rewrite the equation in terms of only cos(x):
5cos²x - 3sin(x)cos(x) + 1 = 0
Solving this quadratic equation in cos(x) might be a bit tricky. We might need to use the quadratic formula:
cos(x) = [3sin(x) ± sqrt((3sin(x))² - 451)] / 10
We can then use the Pythagorean identity sin²x + cos²x = 1 to find the corresponding sine values for the solutions of cos(x).
I hope this helps! Let me know if you have any questions.