How do you use the quadratic formula to solve $3 {csc}^{2} θ - 2 csc θ = 2$ $3csc^2theta-2csctheta=2$ for $0 \leq θ < 360$ $0<=theta<360$ ?

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Answer 1 · 2017-10-19T17:46:45

see below

let $u = csc θ$ $u=csc theta$ then $3 {csc}^{2} θ - 2 csc θ = 2$ $3 csc^2 theta-2csctheta=2$ becomes $3 u^{2} - 2 u = 2$ $3u^2-2u=2$

then if we put everything on one side we have $3 u^{2} - 2 u - 2 = 0$ $3u^2-2u-2=0$ . So let's use the quadratic formula to solve.

Now take $a = 3, b = - 2, and c = - 2$ $a=3,b=-2, and c=-2$ and put it in to the

formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$u = \frac{2 \pm \sqrt{{(- 2)}^{2} - 4 (3) (- 2)}}{2 (3)}$ $u=(2+-sqrt((-2)^2-4(3)(-2)))/(2(3))$

$= \frac{2 \pm \sqrt{28}}{6}$ $=(2+-sqrt(28))/6$

$= \frac{2 \pm 2 \sqrt{7}}{6}$ $=(2+- 2sqrt(7))/6$

$=(cancel2(1+- sqrt(7)))/(cancel(6)3)$

$csctheta=(1+- sqrt(7))/3$

$theta=csc^-1 ((1+- sqrt(7))/3)=sin^-1 (3/(1+- sqrt(7)))$

$theta=sin^-1 (3/(1+ sqrt(7))) or theta = cancel(sin^-1 (3/(1- sqrt(7)))=undef$

$theta=55.37370265+360n^@, or theta =180^@-55.37370265+360n^@$

$theta=55.37370265^@+360n^@, or theta=124.6262974^@+360n^@$

Now let's put in n values (integers) to determine our final answer

$n=0, theta=55.37^@,124.636^@$

Note that if we put in other n-values our answers will be outside of the domain so our solution set is

$S={55.37^@,124.636^@}$

How do you use the quadratic formula to solve 3csc^2theta-2csctheta=23csc2θ−2cscθ=23csc^2theta-2csctheta=2 for 0<=theta<3600≤θ<3600<=theta<360?