3sin-2-sin-cos-4cos-2-0-find-the-values-of-if-lies-between-0-and-360-

Question Number 29354 by NECx last updated on 08/Feb/18

3 \sin^{2} θ - \sin θ \cos θ - 4 \cos^{2} θ = 0

f i n d t h e v a l u e s o f θ i f θ l i e s

b e t w e e n 0 a n d 360

Answered by mrW2 last updated on 08/Feb/18

3 (1 - \cos^{2} θ) - \sin θ \cos θ - 4 \cos^{2} θ = 0

3 - \sin θ \cos θ - 7 \cos^{2} θ = 0

6 - 2 \sin θ \cos θ - 7 (2 \cos^{2} θ) = 0

6 - \sin 2 θ - 7 (1 + \cos 2 θ) = 0

\sin 2 θ + 7 \cos 2 θ = - 1

\sin 2 θ \times \frac{1}{\sqrt{7^{2} + 1^{1}}} + \cos 2 θ \times \frac{7}{\sqrt{7^{2} + 1^{2}}} = - \frac{1}{\sqrt{7^{2} + 1^{2}}}

\sin 2 θ \times \sin α + \cos 2 θ \times \cos α = - \frac{1}{5 \sqrt{2}}

w i t h α = \tan^{- 1} \frac{1}{7}

\Rightarrow \cos (2 θ - α) = - \frac{1}{5 \sqrt{2}}

\Rightarrow 2 θ - α = (2 n + 1) π \pm \cos^{- 1} \frac{1}{5 \sqrt{2}}

\Rightarrow θ = n π + \frac{1}{2} (π + α \pm \cos^{- 1} \frac{1}{5 \sqrt{2}})

\Rightarrow θ = n π + \frac{1}{2} (π + \tan^{- 1} \frac{1}{7} \pm \cos^{- 1} \frac{1}{5 \sqrt{2}})

\Rightarrow θ = 180 n + {\begin{cases} 135 ° \\ 53.1 ° \end{cases}

s o l u t i o n s w i t h i n 0 ° - 360 ° :

\Rightarrow θ = 53.1 °, 135 °, 233.1 °, 315 °

Answered by mrW2 last updated on 08/Feb/18

a n o t h e r w a y :

(3 \sin θ - 4 \cos θ) (\sin θ + \cos θ) = 0

\Rightarrow 3 \sin θ - 4 \cos θ = 0

\Rightarrow \sin θ + \cos θ = 0

\Rightarrow \tan θ = \frac{4}{3}

\Rightarrow \tan θ = - 1

\Rightarrow θ = n π + \tan^{- 1} \frac{4}{3} o r 180 n + 53.1 °

\Rightarrow θ = n π - \frac{π}{4} o r 180 n - 45 °

w i t h i n 0 - 360 ° :

θ = 53.1 °, 135 °, 233.1 °, 315 °

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