Respuesta :
Answer:
θ = [tex]\frac{\pi }{3}[/tex] (60° )
Step-by-step explanation:
Using the identity
sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x
Given
cos²θ - sin²θ = 2 - 5cosθ
cos²θ - (1 - cos²θ) = 2 - 5cosθ
cos²θ - 1 + cos²θ = 2 - 5cosθ
2cos²θ - 1 = 2 - 5cosθ ( subtract 2 - 5cosθ from both sides )
2cos²θ + 5cosθ - 3 = 0 ← in standard form
(cosθ + 3)(2cosθ - 1) = 0 ← in factored form
Equate each factor to zero and solve for θ
cosθ + 3 = 0
cosθ = - 3 ← not possible as - 1 ≤ cosθ ≤ 1
2cosθ - 1 = 0
cosθ = [tex]\frac{1}{2}[/tex] , so
θ = [tex]cos^{-1}[/tex] ([tex]\frac{1}{2}[/tex] ) = [tex]\frac{\pi }{3}[/tex] ( or 60° )
Answer:
[tex] \huge \boxed{ \boxed{\blue{ { \theta = 60}^{ \circ} }}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
- trigonometry
- PEMDAS
let's solve:
- [tex] \sf \: rewrite \: \sin ^{2} ( \theta) \: as \: 1 - \cos ^{2} ( \theta) : \\ \sf \implies \: \cos ^{2} ( \theta) - (1 - \cos ^{2} ( \theta)) = 2 - 5 \cos( \theta) [/tex]
- [tex] \sf \: remove \: parentheses \: and \: change \: its \: sign : \\ \sf \implies \: \cos ^{2} ( \theta) - 1 + \cos ^{2} ( \theta)) = 2 - 5 \cos( \theta) [/tex]
- [tex] \sf \: add : \\ \sf \implies \: 2\cos ^{2} ( \theta) - 1 = 2 - 5 \cos( \theta) [/tex]
- [tex] \sf \: move \: left \: hand \: sides \: expression \: to \: right \: hand \: sides \: : \\ \sf \implies \: 2\cos ^{2} ( \theta) + 5 \cos( \theta) - 1 -2 = 0[/tex]
- [tex] \sf \: rewrite \: 5\cos( \theta) \: as \: 6 \cos( \theta) - \cos( \theta) : \\ \sf \implies \: 2\cos ^{2} ( \theta) + 6 \cos( \theta) - \cos( \theta) - 3 = 0[/tex]
- [tex] \sf \:factor \: out \: 2 \cos( \theta) \: and \: - 1 : \\ \sf \implies \: 2\cos ( \theta)( \cos( \theta) + 3 ) -1( \cos( \theta) + 3) = 0[/tex]
- [tex] \sf \: group: \\ \sf \implies \: (2\cos( \theta) - 1) ( \cos( \theta) + 3 ) = 0[/tex]
- [tex] \sf \: rewrite \: as \: two \: seperate \: equation: \\ \sf \implies \: \begin{cases}2\cos( \theta) - 1 = 0\\ \cos( \theta) + 3 = 0 \end{cases} [/tex]
- [tex]\sf add \: 1 \: to \: the\: first \: equation \: and \: substract \: 3 \: from \: the \: second \: equation: \\ \sf \implies \: \begin{cases}2\cos( \theta) = 1\\ \cos( \theta) = - 3 \end{cases} [/tex]
[tex] \sf the \: second \: eqution \: is \: false \: \\ \sf because \: - 1 \leqslant \cos( \theta) \leqslant 1 \: \\ \sf but \: we \: can \: still \: work \: with \: the \: second \: equation[/tex]
- [tex] \sf substract \: both \: sides \: by \: 2 : \\ \implies\frac{ 2\cos( \theta) }{2} = \frac{1}{2} \\ \implies\cos( \theta) = \frac{1}{2} \\ \therefore \: \theta \: = {60}^{ \circ} [/tex]
