
1) Sin(4x-П/6)=02) sin^2x-2sinxCosx=3cos^2x


Ответы на вопрос

На cosx разделили при условии, что cosx ≠ 0, т.е.



Let's solve both of these trigonometric equations step by step:
- Sin(4x - π/6) = 0
To solve this equation, we'll first isolate sin(4x - π/6) and then find the values of x that satisfy the equation:
sin(4x - π/6) = 0
Now, we can find the values of x:
4x - π/6 = arcsin(0) (Note that arcsin(0) = 0 and π)
4x - π/6 = 0 or 4x - π/6 = π
Solve for x in both cases:
Case 1: 4x - π/6 = 0 4x = π/6 x = π/24
Case 2: 4x - π/6 = π 4x = π + π/6 4x = 7π/6 x = (7π/6) / 4 x = 7π/24
So, the solutions for the equation are x = π/24 and x = 7π/24.
- sin^2x - 2sinx*cosx = 3cos^2x
To solve this equation, we'll use trigonometric identities to simplify and solve for x:
sin^2x - 2sinx*cosx = 3cos^2x
Now, we can use the Pythagorean identity sin^2x + cos^2x = 1:
(1 - cos^2x) - 2sinx*cosx = 3cos^2x
Now, isolate the terms with cos^2x on one side:
1 - cos^2x - 3cos^2x = 2sinx*cosx
Combine like terms:
1 - 4cos^2x = 2sinx*cosx
Now, we can use the identity 4cos^2x = 4(1 - sin^2x):
1 - 4(1 - sin^2x) = 2sinx*cosx
1 - 4 + 4sin^2x = 2sinx*cosx
-3 + 4sin^2x = 2sinx*cosx
Now, let's simplify further:
4sin^2x - 2sinx*cosx - 3 = 0
Now, let's solve this quadratic equation for sinx using the quadratic formula:
sinx = [2cosx ± √(2cosx)^2 - 44(-3)] / (2*4)
sinx = [2cosx ± √(4cosx^2 + 48)] / 8
sinx = [cosx ± √(cosx^2 + 12)] / 4
The solutions for x depend on the values of cosx:
- If cosx = 0, then sinx = ±√12/4 = ±√3/2.
- If cosx ≠ 0, then sinx = (cosx ± √(cosx^2 + 12)) / 4.
So, the solutions to the equation depend on the values of cosx, and you'll need to further analyze the specific values of cosx to find the corresponding values of sinx and x.


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