Solve for ? cos(x)+sin(x)=1

$\cos (x) + \sin (x) = 1$

Square both sides of the equation.

${(\cos (x) + \sin (x))}^{2} = {(1)}^{2}$

Rewrite ${(\cos (x) + \sin (x))}^{2}$ as $(\cos (x) + \sin (x)) (\cos (x) + \sin (x))$ .

$(\cos (x) + \sin (x)) (\cos (x) + \sin (x)) = {(1)}^{2}$

Expand $(\cos (x) + \sin (x)) (\cos (x) + \sin (x))$ using the FOIL Method.

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Apply the distributive property.

$\cos (x) (\cos (x) + \sin (x)) + \sin (x) (\cos (x) + \sin (x)) = {(1)}^{2}$

Apply the distributive property.

$\cos (x) \cos (x) + \cos (x) \sin (x) + \sin (x) (\cos (x) + \sin (x)) = {(1)}^{2}$

Apply the distributive property.

$\cos (x) \cos (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Simplify and combine like terms.

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Simplify each term.

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Multiply $\cos (x) \cos (x)$ .

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Raise $\cos (x)$ to the power of $1$ .

$\cos (x) \cos (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Raise $\cos (x)$ to the power of $1$ .

$\cos (x) \cos (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${\cos (x)}^{1 + 1} + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Add $1$ and $1$ .

$\cos^{2} (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Multiply $\sin (x) \sin (x)$ .

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Raise $\sin (x)$ to the power of $1$ .

$\cos^{2} (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Raise $\sin (x)$ to the power of $1$ .

$\cos^{2} (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin (x) \sin (x) = {(1)}^{2}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\cos^{2} (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + {\sin (x)}^{1 + 1} = {(1)}^{2}$

Add $1$ and $1$ .

$\cos^{2} (x) + \cos (x) \sin (x) + \sin (x) \cos (x) + \sin^{2} (x) = {(1)}^{2}$

Reorder the factors of $\sin (x) \cos (x)$ .

$\cos^{2} (x) + \cos (x) \sin (x) + \cos (x) \sin (x) + \sin^{2} (x) = {(1)}^{2}$

Add $\cos (x) \sin (x)$ and $\cos (x) \sin (x)$ .

$\cos^{2} (x) + 2 \cos (x) \sin (x) + \sin^{2} (x) = {(1)}^{2}$

Move $\sin^{2} (x)$ .

$\cos^{2} (x) + \sin^{2} (x) + 2 \cos (x) \sin (x) = {(1)}^{2}$

Apply pythagorean identity.

$1 + 2 \cos (x) \sin (x) = {(1)}^{2}$

One to any power is one.

$1 + 2 \cos (x) \sin (x) = 1$

Simplify each term.

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Reorder $2 \cos (x)$ and $\sin (x)$ .

$1 + \sin (x) (2 \cos (x)) = 1$

Reorder $\sin (x)$ and $2$ .

$1 + 2 \cdot (\sin (x) \cos (x)) = 1$

Apply the sine double–angle identity.

$1 + \sin (2 x) = 1$

Move all terms not containing $\sin (2 x)$ to the right side of the equation.

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Subtract $1$ from both sides of the equation.

$\sin (2 x) = 1 - 1$

Subtract $1$ from $1$ .

$\sin (2 x) = 0$

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$2 x = \arcsin (0)$

The exact value of $\arcsin (0)$ is $0$ .

$2 x = 0$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Divide $0$ by $2$ .

$x = 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$2 x = π - 0$

Simplify the expression to find the second solution.

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Multiply $- 1$ by $0$ .

$2 x = π + 0$

Add $π$ and $0$ .

$2 x = π$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Find the period.

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The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Solve the equation.

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The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

The period of the $\sin (2 x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = π n, \frac{π}{2} + π n$ , for any integer $n$

Consolidate the answers.

$x = \frac{π n}{2}$ , for any integer $n$

Verify each of the solutions by substituting them into $\cos (x) + \sin (x) = 1$ and solving.

$x = \frac{π}{2} + 2 π n, 2 π n$ , for any integer $n$

Solve for ? cos(x)+sin(x)=1

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