Find All Complex Solutions sin(x)+cos(x)=1

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

Square both sides of the equation.

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

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

Apply pythagorean identity.

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

Simplify each term.

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

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

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

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

Apply the sine double-angle identity.

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

One to any power is one.

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

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

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.

$2 x = \arcsin (0)$

Simplify the right side.

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

$2 x = 0$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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.

$2 x = π - 0$

Solve for $x$ .

Simplify.

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Add $π$ and $0$ .

$2 x = π$

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

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

$\frac{2 x}{2} = \frac{π}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Find the period of $\sin (2 x)$ .

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.

$\frac{2 π}{| 2 |}$

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 $2$ .

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 $n$

Exclude the solutions that do not make $\sin (x) + \cos (x) = 1$ true.

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

Find All Complex Solutions sin(x)+cos(x)=1

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