Trigonometry Examples

Solve for x sec(x)csc(x)=2csc(x)
sec(x)csc(x)=2csc(x)
Step 1
Subtract 2csc(x) from both sides of the equation.
sec(x)csc(x)-2csc(x)=0
Step 2
Factor csc(x) out of sec(x)csc(x)-2csc(x).
Tap for more steps...
Step 2.1
Factor csc(x) out of sec(x)csc(x).
csc(x)sec(x)-2csc(x)=0
Step 2.2
Factor csc(x) out of -2csc(x).
csc(x)sec(x)+csc(x)-2=0
Step 2.3
Factor csc(x) out of csc(x)sec(x)+csc(x)-2.
csc(x)(sec(x)-2)=0
csc(x)(sec(x)-2)=0
Step 3
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
csc(x)=0
sec(x)-2=0
Step 4
Set csc(x) equal to 0 and solve for x.
Tap for more steps...
Step 4.1
Set csc(x) equal to 0.
csc(x)=0
Step 4.2
The range of cosecant is y-1 and y1. Since 0 does not fall in this range, there is no solution.
No solution
No solution
Step 5
Set sec(x)-2 equal to 0 and solve for x.
Tap for more steps...
Step 5.1
Set sec(x)-2 equal to 0.
sec(x)-2=0
Step 5.2
Solve sec(x)-2=0 for x.
Tap for more steps...
Step 5.2.1
Add 2 to both sides of the equation.
sec(x)=2
Step 5.2.2
Take the inverse secant of both sides of the equation to extract x from inside the secant.
x=arcsec(2)
Step 5.2.3
Simplify the right side.
Tap for more steps...
Step 5.2.3.1
The exact value of arcsec(2) is π3.
x=π3
x=π3
Step 5.2.4
The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from 2π to find the solution in the fourth quadrant.
x=2π-π3
Step 5.2.5
Simplify 2π-π3.
Tap for more steps...
Step 5.2.5.1
To write 2π as a fraction with a common denominator, multiply by 33.
x=2π33-π3
Step 5.2.5.2
Combine fractions.
Tap for more steps...
Step 5.2.5.2.1
Combine 2π and 33.
x=2π33-π3
Step 5.2.5.2.2
Combine the numerators over the common denominator.
x=2π3-π3
x=2π3-π3
Step 5.2.5.3
Simplify the numerator.
Tap for more steps...
Step 5.2.5.3.1
Multiply 3 by 2.
x=6π-π3
Step 5.2.5.3.2
Subtract π from 6π.
x=5π3
x=5π3
x=5π3
Step 5.2.6
Find the period of sec(x).
Tap for more steps...
Step 5.2.6.1
The period of the function can be calculated using 2π|b|.
2π|b|
Step 5.2.6.2
Replace b with 1 in the formula for period.
2π|1|
Step 5.2.6.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
2π1
Step 5.2.6.4
Divide 2π by 1.
2π
2π
Step 5.2.7
The period of the sec(x) function is 2π so values will repeat every 2π radians in both directions.
x=π3+2πn,5π3+2πn, for any integer n
x=π3+2πn,5π3+2πn, for any integer n
x=π3+2πn,5π3+2πn, for any integer n
Step 6
The final solution is all the values that make csc(x)(sec(x)-2)=0 true.
x=π3+2πn,5π3+2πn, for any integer n
sec(x)csc(x)=2csc(x)
(
(
)
)
|
|
[
[
]
]
°
°
7
7
8
8
9
9
θ
θ
4
4
5
5
6
6
/
/
^
^
×
×
>
>
π
π
1
1
2
2
3
3
-
-
+
+
÷
÷
<
<
,
,
0
0
.
.
%
%
=
=
 [x2  12  π  xdx ]