Question

M110) solve
csc ( to power 5 ) theta - 4 csc theta + 1 = 0

Answer 1

Answer: 0 is cos^4(theta) = (4 - 1/sin(theta))(1-sin^2(theta))^2.

Step-by-step explanation: To solve the equation csc^5(theta) - 4csc(theta) + 1 = 0, we can use the identity csc^n(theta) = 1/sin^n(theta).

First, we can apply the identity:

1/sin^5(theta) - 4/sin(theta) + 1 = 0

Next, we can factor out sin(theta) from the first two terms:

sin(theta)(1/sin^4(theta) - 4) + 1 = 0

We can then use the identity sin^2(theta) = 1 - cos^2(theta) to rewrite the first term:

sin(theta)(1 - cos^4(theta)/(1-cos^2(theta))^2 - 4) + 1 = 0

We can then divide both sides of the equation by sin(theta), and we get:

1 - cos^4(theta)/(1-cos^2(theta))^2 - 4 + 1/sin(theta) = 0

We can then simplify this equation to get:

cos^4(theta)/(1-cos^2(theta))^2 = 4 - 1/sin(theta)

By using the identity cos^2(theta) = 1-sin^2(theta) we can simplify the equation further:

cos^4(theta) = (4 - 1/sin(theta))(1-sin^2(theta))^2

So to find the solution we can use double angle identities to simplify the right side of the equation.

Therefore, the solution to the equation csc^5(theta) - 4csc(theta) + 1 = 0 is cos^4(theta) = (4 - 1/sin(theta))(1-sin^2(theta))^2.

Answer 2

The equation csc^5 (theta) - 4 csc (theta) + 1 = 0 can be solved by factoring the left-hand side of the equation.

csc^5 (theta) - 4 csc (theta) + 1 = 0

csc (theta) (csc^4 (theta) - 4) + 1 = 0

csc (theta) (csc^2 (theta) - 2)(csc^2 (theta) + 2) + 1 = 0

We can see that the last expression is a polynomial equation of the form (ax+1)(bx^2+c*x+d)=0 where x= csc(theta) , a=1, b=1, c= -2, d=2

Solving for x = 0, we get x=0 which is not a valid solution because csc(theta) is defined only for theta != k*pi where k is an integer.

Solving for x= -1/2 and x= 1/2, we get csc(theta) = -1/2 and csc(theta) = 1/2

To find theta, we can use the reciprocal cosecant function (csc) which is defined as 1/sin(theta)

csc(theta) = -1/2 => sin(theta) = -2

csc(theta) = 1/2 => sin(theta) = 2