Respuesta :
tan²( θ ) - (1 + √3) tan (θ) + √3 = 0
tan²( θ ) - (tan (θ) + √3 tan (θ)) + √3 = 0
tan²( θ ) - tan (θ) - √3 tan (θ) + √3 = 0
tan( θ ) ( tan (θ) - 1) - √3 ( tan (θ) - 1 ) = 0
( tan( θ ) - 1 ) ( tan( θ ) - √3 ) = 0
tan( θ ) - 1 = 0
θ = π/₄
tan( θ ) - √3 = 0
θ = π/₃
so θ = π/₄ and θ = π/₃
tan²( θ ) - (tan (θ) + √3 tan (θ)) + √3 = 0
tan²( θ ) - tan (θ) - √3 tan (θ) + √3 = 0
tan( θ ) ( tan (θ) - 1) - √3 ( tan (θ) - 1 ) = 0
( tan( θ ) - 1 ) ( tan( θ ) - √3 ) = 0
tan( θ ) - 1 = 0
θ = π/₄
tan( θ ) - √3 = 0
θ = π/₃
so θ = π/₄ and θ = π/₃
Answer:
[tex] \huge \boxed{ \red{ \boxed{\begin{cases} \theta = {45}^{ \circ} \\ \theta= {60}^{ \circ} \end{cases} }}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
- trigonometry
- PEMDAS
let's solve:
distribute tan(θ):
=>tan²(θ)-(tan(θ)+√3tan(θ))+√3=0
remove parentheses:
=>tan²(θ)-tan(θ)-√3tan(θ)+√3=0
so this equation is now in standard form i.e ax²+bx+c=0
we can solve by factoring as we solve quadratic equation
factor out tanθ:
=>tan(θ)(tan(θ)-1)-√3tan(θ)+√3=0
factor out -√3:
=>tan(θ)(tan(θ)-1)-√3(tan(θ)-1)=0
group:
=>(tan(θ)-√3)(tan(θ)-1)=0
separate it as two different equation:
[tex] \implies \begin{cases} \tan( \theta) - 1 = 0 \\ \tan( \theta) - \sqrt{3} = 0 \end{cases}[/tex]
add 1 and √3 to both sides to first and second equation respectively:
[tex] \implies \begin{cases} \tan( \theta) - 1 + 1 = 0 + 1 \\ \tan( \theta) - \sqrt{3} + \sqrt{3} = 0 + \sqrt{3} \end{cases} [/tex]
[tex] \implies \begin{cases} \tan( \theta) = 1 \\ \tan( \theta) = \sqrt{3} \end{cases} [/tex]
[tex] \therefore \begin{cases} \theta = {45}^{ \circ} \\ \theta= {60}^{ \circ} \end{cases} [/tex]
