Find all solutions of equation on interval [0,2pi]
1=cot^2x + cscx
Multiply both sides by sin^2(x).
sin^2(x)=cos^2(x)+sin(x)
sin^2(x)=1-sin^2(x)+sin(x)
2sin^2(x)-sin(x)-1=0
(2sin(x)+1)(sin(x)-1)=0
sin(x)=-1/2, x=7pi/6, x=11pi/6
or
sin(x)=1, x=pi/2
Thank you so much!
To find all solutions of the equation 1 = cot^2x + cscx on the interval [0, 2π], we can follow these steps:
Step 1: Rewrite the equation using trigonometric identities
Recall that cot^2x is equivalent to csc^2x - 1. So, we can rewrite the equation as follows:
1 = (csc^2x - 1) + cscx
Step 2: Simplify the equation
Combine like terms to simplify the equation:
csc^2x + cscx - 2 = 0
Step 3: Substitute cscx with 1/sinx
Since cscx is the reciprocal of sinx, we can substitute cscx with 1/sinx:
(1/sinx)^2 + (1/sinx) - 2 = 0
Step 4: Multiply through by sin^2x
To get rid of the denominators, multiply through the equation by sin^2x:
1 + sinx - 2sin^2x = 0
Step 5: Rearrange the equation
Rearrange the equation into a quadratic form:
2sin^2x - sinx - 1 = 0
Step 6: Solve the quadratic equation
This quadratic equation can now be solved using factoring, completing the square, or using the quadratic formula. In this case, it is not easily factorable. Let's solve it using the quadratic formula:
sinx = [ -b ± √(b^2 - 4ac) ] / 2a
where a = 2, b = -1, and c = -1. Plugging in these values, we have:
sinx = [ -(-1) ± √((-1)^2 - 4(2)(-1)) ] / (2)(2)
= [ 1 ± √(1 + 8) ] / 4
= (1 ± √9) / 4
= (1 ± 3) / 4
So, sinx = (1 + 3) / 4 or sinx = (1 - 3) / 4
This gives us two possible values for sinx:
sinx = 1 or sinx = -1/2
Step 7: Find the corresponding values of x
For sinx = 1, the solutions on the interval [0, 2π] are:
x = π/2 and x = 3π/2
For sinx = -1/2, the solutions on the interval [0, 2π] are:
x = 7π/6 and x = 11π/6
Therefore, the solutions to the equation 1 = cot^2x + cscx on the interval [0, 2π] are:
x = π/2, 3π/2, 7π/6, and 11π/6.