Question

Help, Help, Please

Answer 1

Answer:

θ = $\frac{\pi }{3}$ (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 = $\frac{1}{2}$ , so

θ = $cos^{-1}$ ($\frac{1}{2}$ ) = $\frac{\pi }{3}$ ( or 60° )

Answer 2

Answer:

$\huge \boxed{ \boxed{\blue{ { \theta = 60}^{ \circ} }}}$

Step-by-step 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. $\sf substract \: both \: sides \: by \: 2 : \\ \implies\frac{ 2\cos( \theta) }{2} = \frac{1}{2} \\ \implies\cos( \theta) = \frac{1}{2} \\ \therefore \: \theta \: = {60}^{ \circ}$