sin-2-7cos-2-6-find-

Question Number 11781 by tawa last updated on 31/Mar/17

\sin (2 θ) + 7 \cos (2 θ) = 6

find θ

Answered by sma3l2996 last updated on 31/Mar/17

2 s i n (θ) c o s (θ) + 7 {c o s}^{2} (θ) - 7 {s i n}^{2} (θ) = 6

2 s i n (θ) c o s (θ) + 14 {c o s}^{2} (θ) - 7 = 6

2 s i n (θ) c o s (θ) = 13 - 14 {c o s}^{2} (θ)

4 (1 - {c o s}^{2} (θ)) {c o s}^{2} (θ) = 13^{2} + 14^{2} {c o s}^{4} (θ) - 2 \times 13 \times 14 {c o s}^{2} (θ)

4 {c o s}^{2} (θ) - 4 {c o s}^{4} (θ) = 169 + 196 {c o s}^{4} (θ) - 364 {c o s}^{2} (θ)

200 {c o s}^{4} (θ) - 368 {c o s}^{2} (θ) + 169 = 0

200 {c o s}^{4} (θ) - \frac{2 \times 184 \times 10 \sqrt{2}}{10 \sqrt{2}} {c o s}^{2} (θ) + {(\frac{184}{10 \sqrt{2}})}^{2} = {(\frac{92}{5 \sqrt{2}})}^{2} - 169

{(10 \sqrt{2} {c o s}^{2} (θ) - \frac{92}{5 \sqrt{2}})}^{2} = \frac{4232}{25} - 169

10 \sqrt{2} {c o s}^{2} (θ) - \frac{92}{5 \sqrt{2}} = \underline{+} \frac{\sqrt{3387}}{5}

{c o s}^{2} (θ) = (\underline{+} \frac{\sqrt{3387}}{5} + \frac{92}{5 \sqrt{2}}) \frac{1}{10 \sqrt{2}} = \underline{+} \frac{\sqrt{3387}}{50 \sqrt{2}} + \frac{23}{25}

c o s (θ) = \sqrt{\frac{\sqrt{3387}}{50 \sqrt{2}} + \frac{23}{25}}

θ = a c o s (\sqrt{\frac{\sqrt{3387}}{50 \sqrt{2}} + \frac{23}{25}}) + 2 k π

Commented by tawa last updated on 31/Mar/17

God bless you sir

Answered by sandy_suhendra last updated on 31/Mar/17

\sin 2 θ + 7 \cos 2 θ = k \cos (2 θ - α)

= k \cos 2 θ \cos α + k \sin 2 θ \sin α

thus

k \cos α = 7 and k \sin α = 1

{(k \cos α)}^{2} + {(k \sin α)}^{2} = 7^{2} + 1^{2}

k^{2} (\sin^{2} α + \cos^{2} α) = 50

k = \sqrt{50} = 5 \sqrt{2}

\frac{k \sin α}{k \cos α} = \tan α \Rightarrow \tan α = \frac{1}{7} \Rightarrow α = 8.13 °

5 \sqrt{2} \cos (2 θ - 8.13 °) = 6

\cos (2 θ - 8.13 °) = \frac{6}{5 \sqrt{2}}

(1) 2 θ - 8.13 ° = 31.95 ° + p . 360 °

2 θ = 40.08 ° + p . 360 °

θ = 20.04 + p . 180 ° \Rightarrow p = 0, 1, 2, \dots

(2) 2 θ - 8.13 ° = - 31.95 ° + p . 360 °

2 θ = - 23.82 ° + p . 360 °

θ = - 11.91 ° + p . 180 ° \Rightarrow p = 1, 2, 3, \dots

Commented by tawa last updated on 31/Mar/17

God bless you sir .

Answered by ajfour last updated on 31/Mar/17

2 \sin θ \cos θ + 7 (\cos^{2} θ - \sin^{2} θ) = 6

d i v i d i n g b y \cos^{2} θ w e g e t :

2 \tan θ + 7 (1 - \tan^{2} θ) = 6 \sec^{2} θ

2 \tan θ + 7 - 7 \tan^{2} θ = 6 + 6 \tan^{2} θ

13 \tan^{2} θ - 2 \tan θ - 1 = 0

\tan θ = \frac{2 \pm \sqrt{4 + 52}}{26}

θ = \tan^{- 1} (\frac{1 \pm \sqrt{14}}{13}) + n π .

Commented by tawa last updated on 31/Mar/17

God bless you sir .

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