Divide each term in <math><mstyle displaystyle="true"><msqrt><mn>3</mn></msqrt><mi>cot</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>=</mo><mn>1</mn></mstyle></math> by <math><mstyle displaystyle="true"><msqrt><mn>3</mn></msqrt></mstyle></math> .

Simplify the left side.

Cancel the common factor of <math><mstyle displaystyle="true"><msqrt><mn>3</mn></msqrt></mstyle></math> .

Cancel the common factor.

Divide <math><mstyle displaystyle="true"><mi>cot</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> by <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Simplify the right side.

Multiply <math><mstyle displaystyle="true"><mfrac><mrow><mn>1</mn></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></mstyle></math> by <math><mstyle displaystyle="true"><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></mstyle></math> .

Combine and simplify the denominator.

Multiply <math><mstyle displaystyle="true"><mfrac><mrow><mn>1</mn></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></mstyle></math> and <math><mstyle displaystyle="true"><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><msqrt><mn>3</mn></msqrt></mrow></mfrac></mstyle></math> .

Raise <math><mstyle displaystyle="true"><msqrt><mn>3</mn></msqrt></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Raise <math><mstyle displaystyle="true"><msqrt><mn>3</mn></msqrt></mstyle></math> to the power of <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Use the power rule <math><mstyle displaystyle="true"><msup><mrow><mi>a</mi></mrow><mrow><mi>m</mi></mrow></msup><msup><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msup><mo>=</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>m</mi><mo>+</mo><mi>n</mi></mrow></msup></mstyle></math> to combine exponents.

Add <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> and <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Rewrite <math><mstyle displaystyle="true"><msup><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><mn>2</mn></mrow></msup></mstyle></math> as <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> .

Use <math><mstyle displaystyle="true"><mroot><mrow><msup><mrow><mi>a</mi></mrow><mrow><mi>x</mi></mrow></msup></mrow><mrow><mi>n</mi></mrow></mroot><mo>=</mo><msup><mrow><mi>a</mi></mrow><mrow><mfrac><mrow><mi>x</mi></mrow><mrow><mi>n</mi></mrow></mfrac></mrow></msup></mstyle></math> to rewrite <math><mstyle displaystyle="true"><msqrt><mn>3</mn></msqrt></mstyle></math> as <math><mstyle displaystyle="true"><msup><mrow><mn>3</mn></mrow><mrow><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></msup></mstyle></math> .

Apply the power rule and multiply exponents, <math><mstyle displaystyle="true"><msup><mrow><mo>(</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>m</mi></mrow></msup><mo>)</mo></mrow><mrow><mi>n</mi></mrow></msup><mo>=</mo><msup><mrow><mi>a</mi></mrow><mrow><mi>m</mi><mi>n</mi></mrow></msup></mstyle></math> .

Combine <math><mstyle displaystyle="true"><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mstyle></math> and <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Cancel the common factor of <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Cancel the common factor.

Rewrite the expression.

Evaluate the exponent.

Take the inverse cotangent of both sides of the equation to extract <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> from inside the cotangent.

The exact value of <math><mstyle displaystyle="true"><mi>arccot</mi><mrow><mo>(</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt></mrow><mrow><mn>3</mn></mrow></mfrac><mo>)</mo></mrow></mstyle></math> is <math><mstyle displaystyle="true"><mfrac><mrow><mi>π</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> .

The cotangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> to find the solution in the fourth quadrant.

To write <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> as a fraction with a common denominator, multiply by <math><mstyle displaystyle="true"><mfrac><mrow><mn>3</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> .

Combine fractions.

Combine <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> and <math><mstyle displaystyle="true"><mfrac><mrow><mn>3</mn></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> .

Combine the numerators over the common denominator.

Simplify the numerator.

Move <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> to the left of <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> .

Add <math><mstyle displaystyle="true"><mn>3</mn><mi>π</mi></mstyle></math> and <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> .

The period of the function can be calculated using <math><mstyle displaystyle="true"><mfrac><mrow><mi>π</mi></mrow><mrow><mrow><mo>|</mo><mi>b</mi><mo>|</mo></mrow></mrow></mfrac></mstyle></math> .

Replace <math><mstyle displaystyle="true"><mi>b</mi></mstyle></math> with <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> in the formula for period.

The absolute value is the distance between a number and zero. The distance between <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> and <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> is <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Divide <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> by <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

The period of the <math><mstyle displaystyle="true"><mi>cot</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> function is <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> so values will repeat every <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> radians in both directions.

Consolidate the answers.

Do you know how to Solve for x in Radians square root of 3cot(x)=1? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.

Name | two billion one hundred forty-seven million ninety-two thousand forty |
---|

- 2147092040 has 32 divisors, whose sum is
**8695722924** - The reverse of 2147092040 is
**0402907412** - Previous prime number is
**5**

- Is Prime? no
- Number parity even
- Number length 10
- Sum of Digits 29
- Digital Root 2

Name | one billion four hundred thirty-one million nine hundred seventy-two thousand eight hundred forty |
---|

- 1431972840 has 128 divisors, whose sum is
**7773262848** - The reverse of 1431972840 is
**0482791341** - Previous prime number is
**191**

- Is Prime? no
- Number parity even
- Number length 10
- Sum of Digits 39
- Digital Root 3

Name | one billion six hundred seventy-seven million nine hundred sixty-nine thousand one hundred fifty-four |
---|

- 1677969154 has 32 divisors, whose sum is
**2820312000** - The reverse of 1677969154 is
**4519697761** - Previous prime number is
**89**

- Is Prime? no
- Number parity even
- Number length 10
- Sum of Digits 55
- Digital Root 1