Replace the <math><mstyle displaystyle="true"><mn>2</mn><msup><mi>sin</mi><mrow><mn>2</mn></mrow></msup><mrow><mo>(</mo><mi>x</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>x</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>3</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>x</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><mn>3</mn><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"><mn>3</mn><mi>u</mi></mstyle></math> .

Rewrite <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> as <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn><mrow><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><mrow><mo>(</mo><mo>-</mo><mn>3</mn><mi>u</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><mrow><mo>(</mo><mn>2</mn><msup><mrow><mi>u</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>-</mo><mn>3</mn><mi>u</mi><mo>)</mo></mrow><mo>-</mo><mn>1</mn><mrow><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><mn>1</mn><mo>=</mo><mn>2</mn></mstyle></math> and whose sum is <math><mstyle displaystyle="true"><mi>b</mi><mo>=</mo><mo>-</mo><mn>3</mn></mstyle></math> .

Factor <math><mstyle displaystyle="true"><mo>-</mo><mn>3</mn></mstyle></math> out of <math><mstyle displaystyle="true"><mo>-</mo><mn>3</mn><mi>u</mi></mstyle></math> .

Rewrite <math><mstyle displaystyle="true"><mo>-</mo><mn>3</mn></mstyle></math> as <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> plus <math><mstyle displaystyle="true"><mo>-</mo><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> .

Add <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> to 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>x</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>x</mi></mstyle></math> .

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

Take the inverse cosine of both sides of the equation to extract <math><mstyle displaystyle="true"><mi>x</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><mn>1</mn><mo>)</mo></mrow></mstyle></math> is <math><mstyle displaystyle="true"><mn>0</mn></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.

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>cos</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>cos</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.

List all of the solutions.

Consolidate <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi><mi>n</mi></mstyle></math> and <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi><mo>+</mo><mn>2</mn><mi>π</mi><mi>n</mi></mstyle></math> to <math><mstyle displaystyle="true"><mn>2</mn><mi>π</mi><mi>n</mi></mstyle></math> .

Do you know how to Solve for x in Radians 2sin(x)^2+3cos(x)-3=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 eight hundred thirty-seven million four hundred thirty-six thousand two hundred twenty-one |
---|

- 1837436221 has 8 divisors, whose sum is
**1845392120** - The reverse of 1837436221 is
**1226347381** - Previous prime number is
**337**

- Is Prime? no
- Number parity odd
- Number length 10
- Sum of Digits 37
- Digital Root 1

Name | two hundred fifty-one million seven hundred forty-five thousand seven hundred ninety-nine |
---|

- 251745799 has 4 divisors, whose sum is
**251778600** - The reverse of 251745799 is
**997547152** - Previous prime number is
**20549**

- Is Prime? no
- Number parity odd
- Number length 9
- Sum of Digits 49
- Digital Root 4

Name | one billion three hundred fourteen million two hundred eighty-seven thousand nine hundred forty-seven |
---|

- 1314287947 has 8 divisors, whose sum is
**1342804400** - The reverse of 1314287947 is
**7497824131** - Previous prime number is
**811**

- Is Prime? no
- Number parity odd
- Number length 10
- Sum of Digits 46
- Digital Root 1