Question

Help please!!


`\text{solve the following equation : }`

`\sf\tan^(2)(\theta)-(1+√(3))\tan(\theta)+√(3) = 0`
​

Answer 1

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

Answer 2

`\huge \boxed{ \red{ \boxed{\begin{cases} \theta = {45}^( \circ) \\ \theta= {60}^( \circ) \end{cases} }}}`

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:

`\implies \begin{cases} \tan( \theta) - 1 = 0 \\ \tan( \theta) - √(3) = 0 \end{cases}`

add 1 and √3 to both sides to first and second equation respectively:

`\implies \begin{cases} \tan( \theta) - 1 + 1 = 0 + 1 \\ \tan( \theta) - √(3) + √(3) = 0 + √(3) \end{cases}`

`\implies \begin{cases} \tan( \theta) = 1 \\ \tan( \theta) = √(3) \end{cases}`

`\therefore \begin{cases} \theta = {45}^( \circ) \\ \theta= {60}^( \circ) \end{cases}`