Answer:
[tex] \rm \displaystyle \theta = \{ {150}^{ \circ} , {240}^{ \circ} \}[/tex]
Step-by-step explanation:
we would like to solve the following equation:
[tex] \displaystyle 2 \sin( \theta) \tan( \theta) = - 3[/tex]
to do so we need to put the equation in term of a function of a angle therefore rewrite tanθ:
[tex] \displaystyle 2 \sin( \theta) \cdot \frac{ \sin( \theta) }{ \cos( \theta) } = - 3[/tex]
simplify multiplication:
[tex] \displaystyle \frac{ 2\sin^{2} ( \theta) }{ \cos( \theta) } = - 3[/tex]
cross multiplication:
[tex] \displaystyle 2\sin^{2} ( \theta) = - 3 \cos( \theta) [/tex]
isolate -3cosθ to left hand side and change its sign:
[tex] \displaystyle 2\sin^{2} ( \theta) + 3 \cos( \theta) = 0[/tex]
well we can rewrite sin²θ by applying Pythagorean theorem and that yields:
[tex] \rm \displaystyle 2(1 - \cos^{2} ( \theta)) + 3 \cos( \theta) = 0[/tex]
distribute:
[tex] \rm \displaystyle 2 - 2\cos^{2} ( \theta) + 3 \cos( \theta) = 0[/tex]
now notice that our equation ended up with a pattern and that is Quadratic thus by letting cosθ be x to transform the equation:
[tex] \rm \displaystyle 2 - 2 {x}^{2} + 3 x = 0[/tex]
rearrange it to standard form:
[tex] \rm \displaystyle - 2 {x}^{2} + 3 x + 2 = 0[/tex]
divide both sides by -1:
[tex] \rm \displaystyle 2 {x}^{2} - 3 x - 2 = 0[/tex]
well we can factor the quadratic to do so rewrite -3x as -4x+x:
[tex] \rm \displaystyle 2 {x}^{2} - 4 x + x - 2 = 0[/tex]
factor out 2x:
[tex] \rm \displaystyle 2x ({x}^{} - 2)+ x - 2 = 0[/tex]
factor out 1:
[tex] \rm \displaystyle 2x ({x}^{} - 2)+ 1(x - 2 ) = 0[/tex]
group:
[tex] \rm \displaystyle (2x + 1)(x - 2 ) = 0[/tex]
by Zero Product property we obtain:
[tex] \rm \displaystyle \begin{cases} 2x + 1 = 0\\ x - 2 = 0 \end{cases}[/tex]
cancel 1 from the first equation and add 2 to the e equation:
[tex] \rm \displaystyle \begin{cases} 2x = - 1\\ x = 2\end{cases}[/tex]
substitute back:
[tex] \rm \displaystyle \begin{cases} 2 \cos( \theta) = - 1\\ \cos( \theta) = 2\end{cases}[/tex]
since cosθ is only defined for the interval [1,-1] the second equation is false for any value of θ but we can still continue the first equation
[tex] \rm \displaystyle \begin{cases} 2 \cos( \theta) = - 1\\ \cos( \theta) \neq 2\end{cases}[/tex]
divide both sides by 2:
[tex] \rm \displaystyle \cos( \theta) = \frac{ - 1} {2}[/tex]
by unit circle we acquire:
[tex] \begin{cases} \rm \displaystyle \theta = {150}^{ \circ} \\ \theta = {240}^{ \circ} \end{cases}[/tex]
add period of 2nπ:
[tex] \begin{cases} \rm \displaystyle \theta = {150}^{ \circ} + 2n\pi\\ \theta = {240}^{ \circ} + 2n\pi \end{cases}[/tex]
for the interval [0,2π) θ is only defined when n is 0 Thus:
[tex] \begin{cases} \rm \displaystyle \theta = {150}^{ \circ} + 2(0)\pi\\ \theta = {240}^{ \circ} + 2(0)\pi \end{cases}[/tex]
simplify:
[tex] \begin{cases} \rm \displaystyle \theta = {150}^{ \circ} \\ \theta = {240}^{ \circ} \end{cases}[/tex]
hence,
θ is equal to 150° and 240°
