Replace the <math><mstyle displaystyle="true"><mn>2</mn><msup><mi>sin</mi><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mi>t</mi><mo>)</mo></mrow></mstyle></math> with <math><mstyle displaystyle="true"><mn>2</mn><mrow><mo>(</mo><mn>1</mn><mo>-</mo><msup><mi>cos</mi><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mi>t</mi><mo>)</mo></mrow><mo>)</mo></mrow></mstyle></math> based on the <math><mstyle displaystyle="true"><msup><mi>sin</mi><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>+</mo><msup><mi>cos</mi><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mi>x</mi><mo>)</mo></mrow><mo>=</mo><mn>1</mn></mstyle></math> identity.

Apply the distributive property.

Multiply <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> by <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Multiply <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> by <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Subtract <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> from <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Substitute <math><mstyle displaystyle="true"><mi>u</mi></mstyle></math> for <math><mstyle displaystyle="true"><mi>cos</mi><mrow><mo>(</mo><mi>t</mi><mo>)</mo></mrow></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn><msup><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><mi>u</mi><mo>+</mo><mn>1</mn></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn><msup><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msup></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mi>u</mi></mstyle></math> .

Rewrite <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> as <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn><mrow><mo>(</mo><mo>-</mo><mn>1</mn><mo>)</mo></mrow></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mrow><mo>(</mo><mn>2</mn><msup><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>)</mo></mrow><mo>-</mo><mi>u</mi></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mrow><mo>(</mo><mn>2</mn><msup><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mi>u</mi><mo>)</mo></mrow><mo>-</mo><mn>1</mn><mrow><mo>(</mo><mo>-</mo><mn>1</mn><mo>)</mo></mrow></mstyle></math> .

Factor.

Factor by grouping.

For a polynomial of the form <math><mstyle displaystyle="true"><mi>a</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mi>b</mi><mi>x</mi><mo>+</mo><mi>c</mi></mstyle></math> , rewrite the middle term as a sum of two terms whose product is <math><mstyle displaystyle="true"><mi>a</mi><mo>⋅</mo><mi>c</mi><mo>=</mo><mn>2</mn><mo>⋅</mo><mo>-</mo><mn>1</mn><mo>=</mo><mo>-</mo><mn>2</mn></mstyle></math> and whose sum is <math><mstyle displaystyle="true"><mi>b</mi><mo>=</mo><mn>1</mn></mstyle></math> .

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

Rewrite <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> as <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> plus <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math>

Apply the distributive property.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

Factor out the greatest common factor (GCF) from each group.

Factor the polynomial by factoring out the greatest common factor, <math><mstyle displaystyle="true"><mn>2</mn><mi>u</mi><mo>-</mo><mn>1</mn></mstyle></math> .

Remove unnecessary parentheses.

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"><mn>2</mn><mi>u</mi><mo>-</mo><mn>1</mn></mstyle></math> equal to <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

Solve <math><mstyle displaystyle="true"><mn>2</mn><mi>u</mi><mo>-</mo><mn>1</mn><mo>=</mo><mn>0</mn></mstyle></math> for <math><mstyle displaystyle="true"><mi>u</mi></mstyle></math> .

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

Divide each term in <math><mstyle displaystyle="true"><mn>2</mn><mi>u</mi><mo>=</mo><mn>1</mn></mstyle></math> by <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> and simplify.

Divide each term in <math><mstyle displaystyle="true"><mn>2</mn><mi>u</mi><mo>=</mo><mn>1</mn></mstyle></math> by <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Simplify the left side.

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

Cancel the common factor.

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

Set <math><mstyle displaystyle="true"><mi>u</mi><mo>+</mo><mn>1</mn></mstyle></math> equal to <math><mstyle displaystyle="true"><mn>0</mn></mstyle></math> .

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

The final solution is all the values that make <math><mstyle displaystyle="true"><mo>-</mo><mrow><mo>(</mo><mn>2</mn><mi>u</mi><mo>-</mo><mn>1</mn><mo>)</mo></mrow><mrow><mo>(</mo><mi>u</mi><mo>+</mo><mn>1</mn><mo>)</mo></mrow><mo>=</mo><mn>0</mn></mstyle></math> true.

Substitute <math><mstyle displaystyle="true"><mi>cos</mi><mrow><mo>(</mo><mi>t</mi><mo>)</mo></mrow></mstyle></math> for <math><mstyle displaystyle="true"><mi>u</mi></mstyle></math> .

Set up each of the solutions to solve for <math><mstyle displaystyle="true"><mi>t</mi></mstyle></math> .

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

Simplify the right side.

The exact value of <math><mstyle displaystyle="true"><mi>arccos</mi><mrow><mo>(</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</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 cosine 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.

Simplify <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi><mo>-</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>3</mn></mrow></mfrac></mstyle></math> .

To write <math><mstyle displaystyle="true"><mn>2</mn><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"><mn>2</mn><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.

Multiply <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> by <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> .

Subtract <math><mstyle displaystyle="true"><mi>π</mi></mstyle></math> from <math><mstyle displaystyle="true"><mn>6</mn><mi>π</mi></mstyle></math> .

Find the period of <math><mstyle displaystyle="true"><mi>cos</mi><mrow><mo>(</mo><mi>t</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>cos</mi><mrow><mo>(</mo><mi>t</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.

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

Simplify the right side.

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

The cosine function is negative in the second and third 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 third quadrant.

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

Find the period of <math><mstyle displaystyle="true"><mi>cos</mi><mrow><mo>(</mo><mi>t</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>cos</mi><mrow><mo>(</mo><mi>t</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.

List all of the solutions.

Consolidate the answers.

Do you know how to Solve for t in Radians 2sin(t)^2-cos(t)-1=0? 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 five hundred thirty-nine million eight hundred ninety-three thousand thirty-nine |
---|

- 1539893039 has 8 divisors, whose sum is
**1596396672** - The reverse of 1539893039 is
**9303989351** - Previous prime number is
**233**

- Is Prime? no
- Number parity odd
- Number length 10
- Sum of Digits 50
- Digital Root 5

Name | eight hundred eighty-seven million five hundred seventeen thousand eight hundred eighty-seven |
---|

- 887517887 has 4 divisors, whose sum is
**890817480** - The reverse of 887517887 is
**788715788** - Previous prime number is
**269**

- Is Prime? no
- Number parity odd
- Number length 9
- Sum of Digits 59
- Digital Root 5

Name | one billion nine hundred eighty-eight million five hundred twenty-four thousand five hundred thirty |
---|

- 1988524530 has 8 divisors, whose sum is
**3049071084** - The reverse of 1988524530 is
**0354258891** - Previous prime number is
**45**

- Is Prime? no
- Number parity even
- Number length 10
- Sum of Digits 45
- Digital Root 9