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 | two billion twenty-seven million nine hundred seventy-six thousand two hundred sixteen |
---|

- 2027976216 has 128 divisors, whose sum is
**10711505664** - The reverse of 2027976216 is
**6126797202** - Previous prime number is
**37**

- Is Prime? no
- Number parity even
- Number length 10
- Sum of Digits 42
- Digital Root 6

Name | three hundred fifteen million three hundred two thousand eight hundred twenty-seven |
---|

- 315302827 has 8 divisors, whose sum is
**360466400** - The reverse of 315302827 is
**728203513** - Previous prime number is
**4129**

- Is Prime? no
- Number parity odd
- Number length 9
- Sum of Digits 31
- Digital Root 4

Name | one hundred forty-six million nine hundred nineteen thousand fifty |
---|

- 146919050 has 16 divisors, whose sum is
**317345256** - The reverse of 146919050 is
**050919641** - Previous prime number is
**5**

- Is Prime? no
- Number parity even
- Number length 9
- Sum of Digits 35
- Digital Root 8