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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2013 · Mathematics · Trigonometric Equations

If 7 θ= θ- 4 θ, then the general value of θ is

\quad If cos⁡7θ=cos⁡θ−sin⁡4θ,\cos 7 \theta=\cos \theta-\sin 4 \theta, then the general value of θ\theta is

Options

  1. Anπ6,nπ3+(−1)nπ18\frac{n \pi}{6}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}
  2. Bnπ3,nπ3+(−1)nπ18\frac{n \pi}{3}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}
  3. Cnπ4,nπ3±π18\frac{n \pi}{4}, \frac{n \pi}{3} \pm \frac{\pi}{18}
  4. Dnπ4,nπ3+(−1)nπ18\frac{n \pi}{4}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}

Correct answer

D. nπ4,nπ3+(−1)nπ18\frac{n \pi}{4}, \frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}

Step-by-step solution

cos⁡7θ=cos⁡θ−sin⁡4θ⇒sin⁡4θ=cos⁡θ−cos⁡7θ⇒sin⁡4θ=2sin⁡4θsin⁡3θ⇒sin⁡4θ(1−2sin⁡3θ)=0∴sin⁡4θ=0\cos 7 \theta=\cos \theta-\sin 4 \theta\Rightarrow \sin 4 \theta=\cos \theta-\cos 7 \theta\Rightarrow \sin 4 \theta=2 \sin 4 \theta \sin 3 \theta\Rightarrow \sin 4 \theta(1-2 \sin 3 \theta)=0\therefore \sin 4 \theta=0 or sin⁡3θ=12⇒4θ=nπ\sin 3 \theta=\frac{1}{2}\Rightarrow 4 \theta=n \pi or 3θ=nπ+(−1)nπ6⇒θ=nπ43 \theta=n \pi+(-1)^{n} \frac{\pi}{6}\Rightarrow \quad \theta=\frac{n \pi}{4} or nπ3+(−1)nπ18\frac{n \pi}{3}+(-1)^{n} \frac{\pi}{18}
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