2cosxcscsx-4cosx-cscx+2=0
solutions?
Factorize by grouping of terms:
2cos(x)csc(x)-4cos(x)-csc(x)+2 = 0
csc(x)(2cos(x)-1) -2(2cos(x)-1) = 0
(csc(x)-2)(2cos(x)-1) = 0
Therefore
csc(x)-2 = 0 OR 2cos(x)-1 = 0
If csc(x)-2=0, 1/sin(x)=2, sin(x)=1/2
If 2cos(x)-1=0, cos(x)=1/2
The question does not mention if it will require a general solution or all solutions between 0 and 2π. So find the respective solutions accordingly.
Easy
To find the solutions to the equation 2cos(x)csc(x) - 4cos(x) - csc(x) + 2 = 0, we can simplify the equation and then solve for x.
Let's rewrite the equation:
2cos(x)csc(x) - 4cos(x) - csc(x) + 2 = 0
Start by replacing csc(x) with 1/sin(x):
2cos(x)(1/sin(x)) - 4cos(x) - (1/sin(x)) + 2 = 0
Now, let's simplify the equation further:
2cos(x)/sin(x) - 4cos(x) - 1/sin(x) + 2 = 0
To continue, we can get a common denominator by multiplying both sides of the equation by sin(x):
2cos(x) - 4cos(x)sin(x) - 1 + 2sin(x) = 0
Rearrange the terms:
2cos(x) - 4cos(x)sin(x) + 2sin(x) - 1 = 0
Next, let's factor by grouping:
(cos(x) - 1)(2sin(x) - 1) = 0
Now, we set each factor equal to zero to find the solutions:
cos(x) - 1 = 0 or 2sin(x) - 1 = 0
For the first equation, cos(x) - 1 = 0, add 1 to both sides:
cos(x) = 1
This equation is satisfied when x = 0.
For the second equation, 2sin(x) - 1 = 0, add 1 to both sides and then divide by 2:
2sin(x) = 1
sin(x) = 1/2
This equation is satisfied when x = π/6 or x = 5π/6.
Therefore, the solutions to the equation 2cos(x)csc(x) - 4cos(x) - csc(x) + 2 = 0 are x = 0, x = π/6, and x = 5π/6.