Algebra
Calculus
Trigonometry
Matrix
Question
Solve the equation
a={arccsc(242)+2kπ−arccsc(242)+π+2kπ,k∈Z
Alternative Form
a≈{0.23676∘+360∘k179.76324∘+360∘k,k∈Z
Alternative Form
a≈{0.004132+2kπ3.13746+2kπ,k∈Z
Evaluate
csc2(a)−11csc(a)×22=0
Find the domain
csc2(a)−11csc(a)×22=0,a=kπ,k∈Z
Multiply the terms
csc2(a)−242csc(a)=0
Factor the expression
More Steps
Calculate
csc2(a)−242csc(a)
Rewrite the expression
csc(a)csc(a)−csc(a)×242
Factor out csc(a) from the expression
csc(a)(csc(a)−242)
csc(a)(csc(a)−242)=0
Separate the equation into 2 possible cases
csc(a)=0csc(a)−242=0
Solve the equation
More Steps
Evaluate
csc(a)=0
Use the inverse trigonometric function
a=arccsc(0)
Calculate
a∈/R
a∈/Rcsc(a)−242=0
Solve the equation
More Steps
Evaluate
csc(a)−242=0
Move the constant to the right-hand side and change its sign
csc(a)=0+242
Removing 0 doesn't change the value,so remove it from the expression
csc(a)=242
Use the inverse trigonometric function
a=arccsc(242)
Calculate
a=arccsc(242)a=−arccsc(242)+π
Add the period of 2kπ,k∈Z to find all solutions
a=arccsc(242)+2kπ,k∈Za=−arccsc(242)+π+2kπ,k∈Z
Find the union
a={arccsc(242)+2kπ−arccsc(242)+π+2kπ,k∈Z
a∈/Ra={arccsc(242)+2kπ−arccsc(242)+π+2kπ,k∈Z
Find the union
a={arccsc(242)+2kπ−arccsc(242)+π+2kπ,k∈Z
Check if the solution is in the defined range
a={arccsc(242)+2kπ−arccsc(242)+π+2kπ,k∈Z,a=kπ,k∈Z
Solution
a={arccsc(242)+2kπ−arccsc(242)+π+2kπ,k∈Z
Alternative Form
a≈{0.23676∘+360∘k179.76324∘+360∘k,k∈Z
Alternative Form
a≈{0.004132+2kπ3.13746+2kπ,k∈Z
Show Solution
