Question

6sin^2x+5sin(p/2-x)=0 [-5p; -7p/2]

Answer 1

6(1-cos²x)+5cosx=0
6cos²x-5cosx-6=0
cosx=a
6a²-5a-6=0
D=25+144=169
a1=(5-13)/12=-2/3
cosx=-2/3
x=π-arccos2/3+2πn,n∈z
x=-4π+arccos2/3∈[-5π;-7π/2]
a2=(5+13)/12=3/2⇒cosx=1,5>1 нет решения

Answer 2

To solve the equation 6sin^2x + 5sin(p/2-x) = 0 over the interval [-5p, -7p/2], we can follow these steps:

1. Simplify the equation: 6sin^2x + 5sin(p/2-x) = 0

2. Rearrange the terms: 6sin^2x = -5sin(p/2-x)

3. Divide both sides by 6 to isolate sin^2x: sin^2x = -5sin(p/2-x)/6

4. Take the square root of both sides to solve for sin(x): sinx = ±√(-5sin(p/2-x)/6)

5. Since the interval is given as [-5p, -7p/2], we need to find the values of x that satisfy this interval.

6. Substitute the values of x within the given interval into the equation and solve for sin(x): For x = -5p: sin(-5p) = ±√(-5sin(p/2-(-5p))/6)

   For x = -7p/2: sin(-7p/2) = ±√(-5sin(p/2-(-7p/2))/6)

7. Evaluate the expressions on both sides of the equation for the given values of x within the interval to find the solution.

Please note that the solution will depend on the specific value of p provided. If you have a specific value for p, please provide it, and I can assist you further.