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Basic Math Examples
sin(x)=90.4⋅9.8÷4820sin(x)=90.4⋅9.8÷4820
Step 1
Step 1.1
Rewrite the division as a fraction.
sin(x)=90.4⋅9.84820sin(x)=90.4⋅9.84820
Step 1.2
Multiply 90.490.4 by 9.89.8.
sin(x)=885.924820sin(x)=885.924820
Step 1.3
Divide 885.92885.92 by 48204820.
sin(x)=0.18380082sin(x)=0.18380082
sin(x)=0.18380082sin(x)=0.18380082
Step 2
Take the inverse sine of both sides of the equation to extract xx from inside the sine.
x=arcsin(0.18380082)x=arcsin(0.18380082)
Step 3
Step 3.1
Evaluate arcsin(0.18380082)arcsin(0.18380082).
x=0.18485176x=0.18485176
x=0.18485176x=0.18485176
Step 4
The sine 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=(3.14159265)-0.18485176x=(3.14159265)−0.18485176
Step 5
Step 5.1
Remove parentheses.
x=3.14159265-0.18485176x=3.14159265−0.18485176
Step 5.2
Remove parentheses.
x=(3.14159265)-0.18485176x=(3.14159265)−0.18485176
Step 5.3
Subtract 0.184851760.18485176 from 3.141592653.14159265.
x=2.95674088x=2.95674088
x=2.95674088x=2.95674088
Step 6
Step 6.1
The period of the function can be calculated using 2π|b|2π|b|.
2π|b|2π|b|
Step 6.2
Replace bb with 11 in the formula for period.
2π|1|2π|1|
Step 6.3
The absolute value is the distance between a number and zero. The distance between 00 and 11 is 11.
2π12π1
Step 6.4
Divide 2π2π by 11.
2π2π
2π2π
Step 7
The period of the sin(x)sin(x) function is 2π2π so values will repeat every 2π2π radians in both directions.
x=0.18485176+2πn,2.95674088+2πnx=0.18485176+2πn,2.95674088+2πn, for any integer nn