Question

In Exercises 127–130, solve each equation on the interval [0, 2𝝅) by first rewriting the equation in terms of sines or cosines. csc² x + csc x - 2 = 0

Accepted Answer

Start by rewriting the given equation in terms of sine, since $$ \csc x = \frac{1}{\sin x} . $$The equation is $$ \csc^2 x + \csc x - 2 = 0 $$, so substitute to get $$ \frac{1}{\sin^2 x} + \frac{1}{\sin x} - 2 = 0 . To $$simplify, multiply the entire equation by $$ \sin^2 x  ($$noting that $$ \sin x 
eq 0  in $$the domain) to clear the denominators: $$ 1 + \sin x - 2 \sin^2 x = 0 . $$Rewrite the equation in standard quadratic form in terms of $$ \sin x $$: $$ -2 \sin^2 x + \sin x + 1 = 0 . $$For easier handling, multiply both sides by $$ -1  to $$get $$ 2 \sin^2 x - \sin x - 1 = 0 . $$Let $$ y = \sin x . $$Now solve the quadratic equation $$ 2y^2 - y - 1 = 0 $$ using the quadratic formula $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$, where $$ a=2 $$, $$ b=-1 $$, and $$ c=-1 . $$After finding the values of $$ y $$, determine which solutions lie within the range $$ [-1, 1] $$ since $$ \sin x $$ must be in this interval. Then, find all $$ x  in$$ $$ [0, 2\pi) $$ such that $$ \sin x = y .$$

In Exercises 127–130, solve each equation on the interval [0, 2𝝅) by first rewriting the equation in terms of sines or cosines. csc² x + csc x - 2 = 0

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Reciprocal Trigonometric Functions

Solving Quadratic Equations in Trigonometric Form

Solving Trigonometric Equations on a Specific Interval