Precalculus Examples

Solve for x csc(x) = square root of 2
csc(x)=2csc(x)=2
Step 1
Take the inverse cosecant of both sides of the equation to extract xx from inside the cosecant.
x=arccsc(2)x=arccsc(2)
Step 2
Simplify the right side.
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Step 2.1
The exact value of arccsc(2)arccsc(2) is π4π4.
x=π4x=π4
x=π4x=π4
Step 3
The cosecant function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from ππ to find the solution in the second quadrant.
x=π-π4x=ππ4
Step 4
Simplify π-π4ππ4.
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Step 4.1
To write ππ as a fraction with a common denominator, multiply by 4444.
x=π44-π4x=π44π4
Step 4.2
Combine fractions.
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Step 4.2.1
Combine ππ and 4444.
x=π44-π4x=π44π4
Step 4.2.2
Combine the numerators over the common denominator.
x=π4-π4x=π4π4
x=π4-π4x=π4π4
Step 4.3
Simplify the numerator.
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Step 4.3.1
Move 44 to the left of ππ.
x=4π-π4x=4ππ4
Step 4.3.2
Subtract ππ from 4π4π.
x=3π4x=3π4
x=3π4x=3π4
x=3π4x=3π4
Step 5
Find the period of csc(x)csc(x).
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Step 5.1
The period of the function can be calculated using 2π|b|2π|b|.
2π|b|2π|b|
Step 5.2
Replace bb with 11 in the formula for period.
2π|1|2π|1|
Step 5.3
The absolute value is the distance between a number and zero. The distance between 00 and 11 is 11.
2π12π1
Step 5.4
Divide 2π2π by 11.
2π2π
2π2π
Step 6
The period of the csc(x)csc(x) function is 2π2π so values will repeat every 2π2π radians in both directions.
x=π4+2πn,3π4+2πnx=π4+2πn,3π4+2πn, for any integer nn
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 [x2  12  π  xdx ]  x2  12  π  xdx