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Trigonometry Examples
csc(x)(3cot2(x)-1)=0csc(x)(3cot2(x)−1)=0
Step 1
If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.
csc(x)=0csc(x)=0
3cot2(x)-1=03cot2(x)−1=0
Step 2
Step 2.1
Set csc(x)csc(x) equal to 00.
csc(x)=0csc(x)=0
Step 2.2
The range of cosecant is y≤-1y≤−1 and y≥1y≥1. Since 00 does not fall in this range, there is no solution.
No solution
No solution
Step 3
Step 3.1
Set 3cot2(x)-13cot2(x)−1 equal to 00.
3cot2(x)-1=03cot2(x)−1=0
Step 3.2
Solve 3cot2(x)-1=03cot2(x)−1=0 for xx.
Step 3.2.1
Add 11 to both sides of the equation.
3cot2(x)=13cot2(x)=1
Step 3.2.2
Divide each term in 3cot2(x)=13cot2(x)=1 by 33 and simplify.
Step 3.2.2.1
Divide each term in 3cot2(x)=13cot2(x)=1 by 33.
3cot2(x)3=133cot2(x)3=13
Step 3.2.2.2
Simplify the left side.
Step 3.2.2.2.1
Cancel the common factor of 33.
Step 3.2.2.2.1.1
Cancel the common factor.
3cot2(x)3=13
Step 3.2.2.2.1.2
Divide cot2(x) by 1.
cot2(x)=13
cot2(x)=13
cot2(x)=13
cot2(x)=13
Step 3.2.3
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
cot(x)=±√13
Step 3.2.4
Simplify ±√13.
Step 3.2.4.1
Rewrite √13 as √1√3.
cot(x)=±√1√3
Step 3.2.4.2
Any root of 1 is 1.
cot(x)=±1√3
Step 3.2.4.3
Multiply 1√3 by √3√3.
cot(x)=±1√3⋅√3√3
Step 3.2.4.4
Combine and simplify the denominator.
Step 3.2.4.4.1
Multiply 1√3 by √3√3.
cot(x)=±√3√3√3
Step 3.2.4.4.2
Raise √3 to the power of 1.
cot(x)=±√3√31√3
Step 3.2.4.4.3
Raise √3 to the power of 1.
cot(x)=±√3√31√31
Step 3.2.4.4.4
Use the power rule aman=am+n to combine exponents.
cot(x)=±√3√31+1
Step 3.2.4.4.5
Add 1 and 1.
cot(x)=±√3√32
Step 3.2.4.4.6
Rewrite √32 as 3.
Step 3.2.4.4.6.1
Use n√ax=axn to rewrite √3 as 312.
cot(x)=±√3(312)2
Step 3.2.4.4.6.2
Apply the power rule and multiply exponents, (am)n=amn.
cot(x)=±√3312⋅2
Step 3.2.4.4.6.3
Combine 12 and 2.
cot(x)=±√3322
Step 3.2.4.4.6.4
Cancel the common factor of 2.
Step 3.2.4.4.6.4.1
Cancel the common factor.
cot(x)=±√3322
Step 3.2.4.4.6.4.2
Rewrite the expression.
cot(x)=±√331
cot(x)=±√331
Step 3.2.4.4.6.5
Evaluate the exponent.
cot(x)=±√33
cot(x)=±√33
cot(x)=±√33
cot(x)=±√33
Step 3.2.5
The complete solution is the result of both the positive and negative portions of the solution.
Step 3.2.5.1
First, use the positive value of the ± to find the first solution.
cot(x)=√33
Step 3.2.5.2
Next, use the negative value of the ± to find the second solution.
cot(x)=-√33
Step 3.2.5.3
The complete solution is the result of both the positive and negative portions of the solution.
cot(x)=√33,-√33
cot(x)=√33,-√33
Step 3.2.6
Set up each of the solutions to solve for x.
cot(x)=√33
cot(x)=-√33
Step 3.2.7
Solve for x in cot(x)=√33.
Step 3.2.7.1
Take the inverse cotangent of both sides of the equation to extract x from inside the cotangent.
x=arccot(√33)
Step 3.2.7.2
Simplify the right side.
Step 3.2.7.2.1
The exact value of arccot(√33) is π3.
x=π3
x=π3
Step 3.2.7.3
The cotangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from π to find the solution in the fourth quadrant.
x=π+π3
Step 3.2.7.4
Simplify π+π3.
Step 3.2.7.4.1
To write π as a fraction with a common denominator, multiply by 33.
x=π⋅33+π3
Step 3.2.7.4.2
Combine fractions.
Step 3.2.7.4.2.1
Combine π and 33.
x=π⋅33+π3
Step 3.2.7.4.2.2
Combine the numerators over the common denominator.
x=π⋅3+π3
x=π⋅3+π3
Step 3.2.7.4.3
Simplify the numerator.
Step 3.2.7.4.3.1
Move 3 to the left of π.
x=3⋅π+π3
Step 3.2.7.4.3.2
Add 3π and π.
x=4π3
x=4π3
x=4π3
Step 3.2.7.5
Find the period of cot(x).
Step 3.2.7.5.1
The period of the function can be calculated using π|b|.
π|b|
Step 3.2.7.5.2
Replace b with 1 in the formula for period.
π|1|
Step 3.2.7.5.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
π1
Step 3.2.7.5.4
Divide π by 1.
π
π
Step 3.2.7.6
The period of the cot(x) function is π so values will repeat every π radians in both directions.
x=π3+πn,4π3+πn, for any integer n
x=π3+πn,4π3+πn, for any integer n
Step 3.2.8
Solve for x in cot(x)=-√33.
Step 3.2.8.1
Take the inverse cotangent of both sides of the equation to extract x from inside the cotangent.
x=arccot(-√33)
Step 3.2.8.2
Simplify the right side.
Step 3.2.8.2.1
The exact value of arccot(-√33) is 2π3.
x=2π3
x=2π3
Step 3.2.8.3
The cotangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from π to find the solution in the third quadrant.
x=2π3-π
Step 3.2.8.4
Simplify the expression to find the second solution.
Step 3.2.8.4.1
Add 2π to 2π3-π.
x=2π3-π+2π
Step 3.2.8.4.2
The resulting angle of 5π3 is positive and coterminal with 2π3-π.
x=5π3
x=5π3
Step 3.2.8.5
Find the period of cot(x).
Step 3.2.8.5.1
The period of the function can be calculated using π|b|.
π|b|
Step 3.2.8.5.2
Replace b with 1 in the formula for period.
π|1|
Step 3.2.8.5.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
π1
Step 3.2.8.5.4
Divide π by 1.
π
π
Step 3.2.8.6
The period of the cot(x) function is π so values will repeat every π radians in both directions.
x=2π3+πn,5π3+πn, for any integer n
x=2π3+πn,5π3+πn, for any integer n
Step 3.2.9
List all of the solutions.
x=π3+πn,4π3+πn,2π3+πn,5π3+πn, for any integer n
Step 3.2.10
Consolidate the solutions.
Step 3.2.10.1
Consolidate π3+πn and 4π3+πn to π3+πn.
x=π3+πn,2π3+πn,5π3+πn, for any integer n
Step 3.2.10.2
Consolidate 2π3+πn and 5π3+πn to 2π3+πn.
x=π3+πn,2π3+πn, for any integer n
x=π3+πn,2π3+πn, for any integer n
x=π3+πn,2π3+πn, for any integer n
x=π3+πn,2π3+πn, for any integer n
Step 4
The final solution is all the values that make csc(x)(3cot2(x)-1)=0 true.
x=π3+πn,2π3+πn, for any integer n