Help, Help, Please
[tex] \sf find \: the \: value \: of \: \theta \: when \: \theta \: is \: acute\: angle[/tex]
[tex] \sf \cos^{2}(\theta) - \sin^{2}(\theta) =2-5 \cos(\theta) [/tex]
[tex]\text{note:explanation is a must}[/tex]

Answers

Answer 1

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 2

Answer:

[tex] \huge \boxed{ \boxed{\blue{ { \theta = 60}^{ \circ} }}}[/tex]

Step-by-step explanation:

to understand thisyou need to know about:trigonometryPEMDASlet'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]


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Answer:

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Step-by-step explanation:

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