Subtract <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> from both sides of the equation.

Add <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> to both sides of the equation.

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

Apply pythagorean identity.

Factor <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> out of <math><mstyle displaystyle="true"><msup><mi>sec</mi><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> out of <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>+</mo><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>⋅</mo><mo>-</mo><mn>1</mn></mstyle></math> .

If any individual factor on the left side of the equation is equal to <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> , the entire expression will be equal to <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

Set <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> equal to <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

The range of secant is <math><mstyle displaystyle="true"><mi>y</mi><mo>≤</mo><mo>-</mo><mn>1</mn></mstyle></math> and <math><mstyle displaystyle="true"><mi>y</mi><mo>≥</mo><mn>1</mn></mstyle></math> . Since <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> does not fall in this range, there is no solution.

No solution

No solution

Set <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>-</mo><mn>1</mn></mstyle></math> equal to <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

Solve <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn></mstyle></math> for <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> .

Add <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> to both sides of the equation.

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

Simplify the right side.

The exact value of <math><mstyle displaystyle="true"><mi>arcsec</mi><mrow><mo>(</mo><mn>1</mn><mo>)</mo></mrow></mstyle></math> is <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi></mstyle></math> to find the solution in the fourth quadrant.

Subtract <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> from <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi></mstyle></math> .

Find the period of <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> .

The period of the function can be calculated using <math><mstyle displaystyle="true"><mfrac><mrow><mn>2</mn><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"><mn>2</mn><mi>π</mi></mstyle></math> by <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

The period of the <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow></mstyle></math> function is <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi></mstyle></math> so values will repeat every <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi></mstyle></math> radians in both directions.

The final solution is all the values that make <math><mstyle displaystyle="true"><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mrow><mo>(</mo><mi>sec</mi><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>-</mo><mn>1</mn><mo>)</mo></mrow><mo>=</mo><mn>0</mn></mstyle></math> true.

Consolidate the answers.

Do you know how to Solve for x in Radians tan(x)^2=sec(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 | one billion fifty-eight million six hundred twenty-two thousand four hundred sixty-five |
---|

- 1058622465 has 16 divisors, whose sum is
**1703590848** - The reverse of 1058622465 is
**5642268501** - Previous prime number is
**173**

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

Name | one billion thirteen million six hundred forty-six thousand four hundred twenty-five |
---|

- 1013646425 has 16 divisors, whose sum is
**1189693440** - The reverse of 1013646425 is
**5246463101** - Previous prime number is
**11**

- Is Prime? no
- Number parity odd
- Number length 10
- Sum of Digits 32
- Digital Root 5

Name | five hundred twenty-nine million seven hundred seventy-five thousand seven hundred fifty-three |
---|

- 529775753 has 8 divisors, whose sum is
**575578248** - The reverse of 529775753 is
**357577925** - Previous prime number is
**113**

- Is Prime? no
- Number parity odd
- Number length 9
- Sum of Digits 50
- Digital Root 5