Trigonometry Examples

Solve for ? cos(2x)+2cos(x)^2=0
cos(2x)+2cos2(x)=0cos(2x)+2cos2(x)=0
Step 1
Simplify the left side of the equation.
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Step 1.1
Use the double-angle identity to transform cos(2x)cos(2x) to 2cos2(x)-12cos2(x)1.
2cos2(x)-1+2cos2(x)=02cos2(x)1+2cos2(x)=0
Step 1.2
Add 2cos2(x)2cos2(x) and 2cos2(x)2cos2(x).
-1+4cos2(x)=01+4cos2(x)=0
-1+4cos2(x)=01+4cos2(x)=0
Step 2
Factor -1+4cos2(x)1+4cos2(x).
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Step 2.1
Rewrite 4cos2(x)4cos2(x) as (2cos(x))2(2cos(x))2.
-1+(2cos(x))2=01+(2cos(x))2=0
Step 2.2
Rewrite 11 as 1212.
-12+(2cos(x))2=012+(2cos(x))2=0
Step 2.3
Reorder -1212 and (2cos(x))2(2cos(x))2.
(2cos(x))2-12=0(2cos(x))212=0
Step 2.4
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b)a2b2=(a+b)(ab) where a=2cos(x)a=2cos(x) and b=1b=1.
(2cos(x)+1)(2cos(x)-1)=0(2cos(x)+1)(2cos(x)1)=0
(2cos(x)+1)(2cos(x)-1)=0(2cos(x)+1)(2cos(x)1)=0
Step 3
If any individual factor on the left side of the equation is equal to 00, the entire expression will be equal to 00.
2cos(x)+1=02cos(x)+1=0
2cos(x)-1=02cos(x)1=0
Step 4
Set 2cos(x)+12cos(x)+1 equal to 00 and solve for xx.
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Step 4.1
Set 2cos(x)+12cos(x)+1 equal to 00.
2cos(x)+1=02cos(x)+1=0
Step 4.2
Solve 2cos(x)+1=02cos(x)+1=0 for xx.
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Step 4.2.1
Subtract 11 from both sides of the equation.
2cos(x)=-12cos(x)=1
Step 4.2.2
Divide each term in 2cos(x)=-12cos(x)=1 by 22 and simplify.
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Step 4.2.2.1
Divide each term in 2cos(x)=-12cos(x)=1 by 22.
2cos(x)2=-122cos(x)2=12
Step 4.2.2.2
Simplify the left side.
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Step 4.2.2.2.1
Cancel the common factor of 22.
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Step 4.2.2.2.1.1
Cancel the common factor.
2cos(x)2=-12
Step 4.2.2.2.1.2
Divide cos(x) by 1.
cos(x)=-12
cos(x)=-12
cos(x)=-12
Step 4.2.2.3
Simplify the right side.
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Step 4.2.2.3.1
Move the negative in front of the fraction.
cos(x)=-12
cos(x)=-12
cos(x)=-12
Step 4.2.3
Take the inverse cosine of both sides of the equation to extract x from inside the cosine.
x=arccos(-12)
Step 4.2.4
Simplify the right side.
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Step 4.2.4.1
The exact value of arccos(-12) is 2π3.
x=2π3
x=2π3
Step 4.2.5
The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from 2π to find the solution in the third quadrant.
x=2π-2π3
Step 4.2.6
Simplify 2π-2π3.
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Step 4.2.6.1
To write 2π as a fraction with a common denominator, multiply by 33.
x=2π33-2π3
Step 4.2.6.2
Combine fractions.
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Step 4.2.6.2.1
Combine 2π and 33.
x=2π33-2π3
Step 4.2.6.2.2
Combine the numerators over the common denominator.
x=2π3-2π3
x=2π3-2π3
Step 4.2.6.3
Simplify the numerator.
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Step 4.2.6.3.1
Multiply 3 by 2.
x=6π-2π3
Step 4.2.6.3.2
Subtract 2π from 6π.
x=4π3
x=4π3
x=4π3
Step 4.2.7
Find the period of cos(x).
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Step 4.2.7.1
The period of the function can be calculated using 2π|b|.
2π|b|
Step 4.2.7.2
Replace b with 1 in the formula for period.
2π|1|
Step 4.2.7.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
2π1
Step 4.2.7.4
Divide 2π by 1.
2π
2π
Step 4.2.8
The period of the cos(x) function is 2π so values will repeat every 2π radians in both directions.
x=2π3+2πn,4π3+2πn, for any integer n
x=2π3+2πn,4π3+2πn, for any integer n
x=2π3+2πn,4π3+2πn, for any integer n
Step 5
Set 2cos(x)-1 equal to 0 and solve for x.
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Step 5.1
Set 2cos(x)-1 equal to 0.
2cos(x)-1=0
Step 5.2
Solve 2cos(x)-1=0 for x.
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Step 5.2.1
Add 1 to both sides of the equation.
2cos(x)=1
Step 5.2.2
Divide each term in 2cos(x)=1 by 2 and simplify.
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Step 5.2.2.1
Divide each term in 2cos(x)=1 by 2.
2cos(x)2=12
Step 5.2.2.2
Simplify the left side.
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Step 5.2.2.2.1
Cancel the common factor of 2.
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Step 5.2.2.2.1.1
Cancel the common factor.
2cos(x)2=12
Step 5.2.2.2.1.2
Divide cos(x) by 1.
cos(x)=12
cos(x)=12
cos(x)=12
cos(x)=12
Step 5.2.3
Take the inverse cosine of both sides of the equation to extract x from inside the cosine.
x=arccos(12)
Step 5.2.4
Simplify the right side.
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Step 5.2.4.1
The exact value of arccos(12) is π3.
x=π3
x=π3
Step 5.2.5
The cosine 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 5.2.6
Simplify 2π-π3.
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Step 5.2.6.1
To write 2π as a fraction with a common denominator, multiply by 33.
x=2π33-π3
Step 5.2.6.2
Combine fractions.
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Step 5.2.6.2.1
Combine 2π and 33.
x=2π33-π3
Step 5.2.6.2.2
Combine the numerators over the common denominator.
x=2π3-π3
x=2π3-π3
Step 5.2.6.3
Simplify the numerator.
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Step 5.2.6.3.1
Multiply 3 by 2.
x=6π-π3
Step 5.2.6.3.2
Subtract π from 6π.
x=5π3
x=5π3
x=5π3
Step 5.2.7
Find the period of cos(x).
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Step 5.2.7.1
The period of the function can be calculated using 2π|b|.
2π|b|
Step 5.2.7.2
Replace b with 1 in the formula for period.
2π|1|
Step 5.2.7.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
2π1
Step 5.2.7.4
Divide 2π by 1.
2π
2π
Step 5.2.8
The period of the cos(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 (2cos(x)+1)(2cos(x)-1)=0 true.
x=2π3+2πn,4π3+2πn,π3+2πn,5π3+2πn, for any integer n
Step 7
Consolidate the answers.
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Step 7.1
Consolidate 2π3+2πn and 5π3+2πn to 2π3+πn.
x=2π3+πn,4π3+2πn,π3+2πn, for any integer n
Step 7.2
Consolidate 4π3+2πn and π3+2πn to π3+πn.
x=2π3+πn,π3+πn, for any integer n
x=2π3+πn,π3+πn, for any integer n
 [x2  12  π  xdx ]