Solve for x sin(x)+csc(x)=2

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

Rewrite $\csc (x)$ in terms of sines and cosines.

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

Substitute $u$ for $\sin (x)$ .

$(u) + \frac{1}{u} = 2$

Find the LCD of the terms in the equation.

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Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, u, 1$

Since $1, u, 1$ contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1$ then find LCM for the variable part $u^{1}$ .

The LCM is the smallest number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $u^{1}$ is $u$ itself.

$u^{1} = u$

$u$ occurs $1$ time.

The LCM of $u^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$u$

Multiply each term by $u$ and simplify.

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Multiply each term in $u + \frac{1}{u} = 2$ by $u$ in order to remove all the denominators from the equation.

$u \cdot u + \frac{1}{u} \cdot u = 2 \cdot u$

Simplify each term.

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Multiply $u$ by $u$ .

$u^{2} + \frac{1}{u} \cdot u = 2 \cdot u$

Cancel the common factor of $u$ .

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Cancel the common factor.

$u^{2} + \frac{1}{u} \cdot u = 2 \cdot u$

Rewrite the expression.

$u^{2} + 1 = 2 \cdot u$

$u^{2} + 1 = 2 u$

Subtract $2 u$ from both sides of the equation.

$u^{2} + 1 - 2 u = 0$

Factor using the perfect square rule.

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Rearrange terms.

$u^{2} - 2 u + 1 = 0$

Rewrite $1$ as $1^{2}$ .

$u^{2} - 2 u + 1^{2} = 0$

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot u \cdot - 1$

Simplify.

$2 a b = - 2 u$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = - 1$ .

${(u - 1)}^{2} = 0$

Set $u - 1$ equal to $0$ and solve for $u$ .

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Set the factor equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

The solution is the result of $u - 1 = 0$ .

$u = 1$

Substitute $\sin (x)$ for $u$ .

$\sin (x) = 1$

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

$x = \arcsin (1)$

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

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.

$x = π - \frac{π}{2}$

Simplify $π - \frac{π}{2}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{π}{1} \cdot \frac{2}{2} - \frac{π}{2}$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{π \cdot 2}{1 \cdot 2} - \frac{π}{2}$

Multiply $2$ by $1$ .

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

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Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

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 $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Solve the equation.

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The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

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

Solve for x sin(x)+csc(x)=2

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