Question

Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Answer 1

$\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ \mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ \mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} \mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} \end{array}$


$\large\begin{array}{l} \textsf{Using the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ \mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ \mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ \mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}$

$\large\begin{array}{l} \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ \mathsf{\Delta=(4.8)^2}\\\\\\ \mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ \mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} \end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ \mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Both are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ \begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or }~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse tangent function:}\\\\ \begin{array}{rcl} \mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or }~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ \mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or }~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ \textsf{where k in an integer.} \end{array}$

__________


$\large\begin{array}{l} \textsf{Now, restrict x values to the interval }\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ \begin{array}{rcl} \mathsf{x=-tan^{-1}(3.4)$


$\large\begin{array}{l} \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} \mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or }~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} \end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{ is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} \mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or }~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} \end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} \end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}$


$\large\begin{array}{l} \textsf{Solution set:}\\\\ \mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}} \end{array}$


If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2069153


$\large\textsf{I hope it helps.}$


Tags: trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function