Trigonometry Examples

Solve over the Interval sin(x)=-( square root of 5)/5 , (3pi)/2<x<2pi
,
Step 1
Take the inverse sine of both sides of the equation to extract from inside the sine.
Step 2
Simplify the right side.
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Step 2.1
Evaluate .
Step 3
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from , to find a reference angle. Next, add this reference angle to to find the solution in the third quadrant.
Step 4
Simplify the expression to find the second solution.
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Step 4.1
Subtract from .
Step 4.2
The resulting angle of is positive, less than , and coterminal with .
Step 5
Find the period of .
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Step 5.1
The period of the function can be calculated using .
Step 5.2
Replace with in the formula for period.
Step 5.3
The absolute value is the distance between a number and zero. The distance between and is .
Step 5.4
Divide by .
Step 6
Add to every negative angle to get positive angles.
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Step 6.1
Add to to find the positive angle.
Step 6.2
Subtract from .
Step 6.3
List the new angles.
Step 7
The period of the function is so values will repeat every radians in both directions.
, for any integer
Step 8
Plug in for and simplify to see if the solution is contained in .
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Step 8.1
Plug in for .
Step 8.2
Simplify.
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Step 8.2.1
Multiply .
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Step 8.2.1.1
Multiply by .
Step 8.2.1.2
Multiply by .
Step 8.2.2
Add and .
Step 8.3
The interval contains .