Trigonometry Examples
2cos(x)-√3=02cos(x)−√3=0
Step 1
Add √3√3 to both sides of the equation.
2cos(x)=√32cos(x)=√3
Step 2
Step 2.1
Divide each term in 2cos(x)=√32cos(x)=√3 by 22.
2cos(x)2=√322cos(x)2=√32
Step 2.2
Simplify the left side.
Step 2.2.1
Cancel the common factor of 22.
Step 2.2.1.1
Cancel the common factor.
2cos(x)2=√322cos(x)2=√32
Step 2.2.1.2
Divide cos(x)cos(x) by 11.
cos(x)=√32cos(x)=√32
cos(x)=√32cos(x)=√32
cos(x)=√32cos(x)=√32
cos(x)=√32cos(x)=√32
Step 3
Take the inverse cosine of both sides of the equation to extract xx from inside the cosine.
x=arccos(√32)x=arccos(√32)
Step 4
Step 4.1
The exact value of arccos(√32)arccos(√32) is π6π6.
x=π6x=π6
x=π6x=π6
Step 5
The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from 2π2π to find the solution in the fourth quadrant.
x=2π-π6x=2π−π6
Step 6
Step 6.1
To write 2π2π as a fraction with a common denominator, multiply by 6666.
x=2π⋅66-π6x=2π⋅66−π6
Step 6.2
Combine fractions.
Step 6.2.1
Combine 2π2π and 6666.
x=2π⋅66-π6x=2π⋅66−π6
Step 6.2.2
Combine the numerators over the common denominator.
x=2π⋅6-π6x=2π⋅6−π6
x=2π⋅6-π6x=2π⋅6−π6
Step 6.3
Simplify the numerator.
Step 6.3.1
Multiply 66 by 22.
x=12π-π6x=12π−π6
Step 6.3.2
Subtract ππ from 12π12π.
x=11π6x=11π6
x=11π6x=11π6
x=11π6x=11π6
Step 7
Step 7.1
The period of the function can be calculated using 2π|b|2π|b|.
2π|b|2π|b|
Step 7.2
Replace bb with 11 in the formula for period.
2π|1|2π|1|
Step 7.3
The absolute value is the distance between a number and zero. The distance between 00 and 11 is 11.
2π12π1
Step 7.4
Divide 2π2π by 11.
2π2π
2π2π
Step 8
The period of the cos(x)cos(x) function is 2π2π so values will repeat every 2π2π radians in both directions.
x=π6+2πn,11π6+2πnx=π6+2πn,11π6+2πn, for any integer nn