Question

Help, Help, Please


`\sf find \: the \: value \: of \: \theta \: when \: \theta \: is \: acute\: angle`

`\sf \cos^(2)(\theta) - \sin^(2)(\theta) =2-5 \cos(\theta)`

`\text{note:explanation is a must}`
​

Answer 1

θ =
`(\pi )/(3)` (60° )

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 =
`(1)/(2)` , so

θ =
`cos^(-1)` (
`(1)/(2)` ) =
`(\pi )/(3)` ( or 60° )

Answer 2

`\huge \boxed{ \boxed{\blue{ { \theta = 60}^( \circ) }}}`

Explanation:

to understand this

you need to know about:

• trigonometry
• PEMDAS

let's solve:

1. 
   `\sf \: rewrite \: \sin ^(2) ( \theta) \: as \: 1 - \cos ^(2) ( \theta) : \\ \sf \implies \: \cos ^(2) ( \theta) - (1 - \cos ^(2) ( \theta)) = 2 - 5 \cos( \theta)`
2. 
   `\sf \: remove \: parentheses \: and \: change \: its \: sign : \\ \sf \implies \: \cos ^(2) ( \theta) - 1 + \cos ^(2) ( \theta)) = 2 - 5 \cos( \theta)`
3. 
   `\sf \: add : \\ \sf \implies \: 2\cos ^(2) ( \theta) - 1 = 2 - 5 \cos( \theta)`
4. 
   `\sf \: move \: left \: hand \: sides \: expression \: to \: right \: hand \: sides \: : \\ \sf \implies \: 2\cos ^(2) ( \theta) + 5 \cos( \theta) - 1 -2 = 0`
5. 
   `\sf \: rewrite \: 5\cos( \theta) \: as \: 6 \cos( \theta) - \cos( \theta) : \\ \sf \implies \: 2\cos ^(2) ( \theta) + 6 \cos( \theta) - \cos( \theta) - 3 = 0`
6. 
   `\sf \:factor \: out \: 2 \cos( \theta) \: and \: - 1 : \\ \sf \implies \: 2\cos ( \theta)( \cos( \theta) + 3 ) -1( \cos( \theta) + 3) = 0`
7. 
   `\sf \: group: \\ \sf \implies \: (2\cos( \theta) - 1) ( \cos( \theta) + 3 ) = 0`
8. 
   `\sf \: rewrite \: as \: two \: seperate \: equation: \\ \sf \implies \: \begin{cases}2\cos( \theta) - 1 = 0\\ \cos( \theta) + 3 = 0 \end{cases}`
9. 
   `\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}`

`\sf the \: second \: eqution \: is \: false \: \\ \sf because \: - 1 \leqslant \cos( \theta) \leqslant 1 \: \\ \sf but \: we \: can \: still \: work \: with \: the \: second \: equation`

1. 
   `\sf substract \: both \: sides \: by \: 2 : \\ \implies( 2\cos( \theta) )/(2) = (1)/(2) \\ \implies\cos( \theta) = (1)/(2) \\ \therefore \: \theta \: = {60}^( \circ)`