Solve:
3csc^2x - 5cscx - 2 = 0
sec^2x + 5secx + 6 = 0
interval : [0, 2pi]
To solve the equation 3csc^2x - 5cscx - 2 = 0, we can use a substitution to simplify the equation.
Let's substitute cscx with 1/sinx:
3(1/sinx)^2 - 5(1/sinx) - 2 = 0
Simplifying this equation, we get:
3/sin^2x - 5/sinx - 2 = 0
Multiplying through by sin^2x to clear the fractions:
3 - 5sinx - 2sin^2x = 0
Rearranging the terms:
2sin^2x + 5sinx - 3 = 0
Now, let's solve the quadratic equation 2sin^2x + 5sinx - 3 = 0:
We can factor this quadratic equation:
(2sinx - 1)(sinx + 3) = 0
Setting each factor equal to zero:
2sinx - 1 = 0 or sinx + 3 = 0
Solving these two equations:
2sinx = 1 or sinx = -3
For the equation 2sinx = 1, dividing both sides by 2:
sinx = 1/2
From the unit circle, we know that at x = π/6 and x = 5π/6, sinx = 1/2.
For the equation sinx = -3, there are no solutions. Since the sine function has a maximum value of 1 and a minimum value of -1, it is not possible for sinx to equal -3.
Therefore, the solutions to the equation 3csc^2x - 5cscx - 2 = 0 are:
x = π/6 + 2kπ or x = 5π/6 + 2kπ (where k is an integer)
Now let's solve the equation sec^2x + 5secx + 6 = 0.
Substitute cosx with 1/secx:
(1/secx)^2 + 5(1/secx) + 6 = 0
Simplifying the equation, we get:
1 + 5secx + 6sec^2x = 0
Multiplying through by sec^2x to clear the fractions:
sec^2x + 5secx + 6 = 0
This quadratic equation can be factored:
(secx + 2)(secx + 3) = 0
Setting each factor equal to zero:
secx + 2 = 0 or secx + 3 = 0
Solving these equations:
secx = -2 or secx = -3
The secant function takes on the value of -1 at π, so it is not possible for secx to equal -2 or -3 because the secant function does not have values between -1 and 1.
Therefore, the equation sec^2x + 5secx + 6 = 0 has no solutions.
In the interval [0, 2π], the solutions to the equation 3csc^2x - 5cscx - 2 = 0 are:
x = π/6 and x = 5π/6
To solve the equation 3csc^2x - 5cscx - 2 = 0, we can first substitute the value of cscx as 1/sinx.
The equation becomes:
3(1/sinx)^2 - 5(1/sinx) - 2 = 0
Simplifying further:
3/sin^2x - 5/sinx - 2 = 0
Multiplying through by sin^2x, we get:
3 - 5sinx - 2sin^2x = 0
Rearranging the terms:
2sin^2x + 5sinx - 3 = 0
Now, let's solve this quadratic equation.
We can either factor or use the quadratic formula to find the solutions.
If we try to factor the equation, it becomes a bit more complex since the coefficients are not very neat. So, let's use the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, the coefficients are:
a = 2, b = 5, c = -3
Substituting these values into the quadratic formula:
x = (-5 ± √(5^2 - 4(2)(-3))) / (2(2))
Simplifying further:
x = (-5 ± √(25 + 24)) / 4
x = (-5 ± √(49)) / 4
x = (-5 ± 7) / 4
So, we have two possible values for x:
x1 = (-5 + 7) / 4 = 2/4 = 1/2
x2 = (-5 - 7) / 4 = -12/4 = -3
Now, we need to check if these values fall within the given interval [0, 2pi].
For x = 1/2:
Since 0 ≤ sin^-1(1/2) ≤ 2π,
x = sin^-1(1/2)
For x = -3:
Since -π ≤ sin^-1(-3) ≤ π,
x = sin^-1(-3) is not within the given interval.
Therefore, the solution to the equation 3csc^2x - 5cscx - 2 = 0 in the interval [0, 2π] is x = sin^-1(1/2) or x = π/6.