Solve the trignometric equation and find ALL solutions
$f (x) = (3 cos x + 7) (- 2 sin x - 1) = 0$

no solutions

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{7 π}{6} + 2 n

π, \frac{11 π}{6} + 2 n

π

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{3} + 2 n

π, \frac{2 π}{3} + 2 n

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{7}{3}, - \frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{7 π}{6} + 2 n$ $π, \frac{11 π}{6} + 2 n$ $π$
Given, $f (x) = (3 cos x + 7) (- 2 sin x - 1)$
$\Rightarrow (3 cos x + 7) (- 2 sin x - 1) = 0$
$\Rightarrow 3 cos x + 7 = 0$ or $- 2 sin x - 1 = 0$
$\Rightarrow cos x = \frac{- 7}{3}$ or $sin x = \frac{- 1}{2}$
$\Rightarrow - 1 \leq cos x \leq 1$
Thus $cos x = \frac{- 7}{3}$ is rejected
$\Rightarrow sin x = \frac{- 1}{2}$
Value of $sin x$ is negative in $3^{r d}$ and $4^{t h}$ quadrant.
Therefore, $sin (π + \frac{π}{6}) = sin \frac{7 π}{6} = - \frac{1}{2}$
$\Rightarrow sin (\frac{3 π}{2} + \frac{π}{3}) = sin \frac{11 π}{6} = - \frac{1}{2}$
$\Rightarrow x = \frac{7 π}{6} + 2 n π, \frac{11 π}{6} + 2 n π$

Suggest Corrections

Similar questions

| 2 sin θ - cosec θ | \geq 1

θ

n \in Z

n \prod n = 1 c o s \frac{x}{2^{n}} = \frac{sin x}{2^{n} sin (\frac{x}{2^{n}})}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} {lim}_{n \to \infty} \sum_{n = 1}^{n} \frac{1}{2^{n}} tan (\frac{x}{2^{n}}), & x \in (0, π) - {\frac{π}{2}} \frac{2}{π} & x = \frac{π}{2} \end{matrix}

f (x) = lim n \to \infty {\frac{x}{1 + {3 {tan}^{2} x}^{n}}}

f (x)

π

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} 0, & - \frac{π}{2} < x < - \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = - \frac{π}{6} \end{matrix} \begin{matrix} x, & - \frac{π}{6} < x < \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = \frac{π}{6} \end{matrix} \begin{matrix} 0, & \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{matrix}

lim n \to \infty \frac{π}{6 n} [{sec}^{2} (\frac{π}{6 n}) + {sec}^{2} (2 \cdot \frac{π}{6 n}) + \dots + {sec}^{2} ((n - 1) \frac{π}{6 n}) + \frac{4}{3}]

\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{(2 n + 1) (2 n + 3)} = \frac{n}{3 (2 n + 3)}

Join BYJU'S Learning Program