The number of solutions of the equation cot2 sin ⁡ x + 3=1 in 0,3 π is equal to

Question

The number of solutions of the equation cot2 sin ⁡ x + 3=1 in 0,3 π is equal to (A) 2 (B) 4 (C) 6 (D) 8

Tardigrade Admin · Accepted Answer

cot2 sin â¡ x + 3=1=cot2 â¡ (π /4) ⇒ sin x+3=nπ ±(π /4) Also, 2≤ sin x+3≤ 4 ⇒ sin x+3=π -(π /4),π +(π /4) ⇒ sin x=(3 π /4)-3 or (5 π /4)-3 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-gygpv4rcim6xtsqi.jpg" />

Q. The number of solutions of the equation $co t^{2} s in x + 3 = 1$ in $0, 3 π$ is equal to

$2$

$4$

$6$

$8$

$co t^{2} s in x + 3 = 1 = co t^{2} \frac{π}{4}$
$\Rightarrow s in x + 3 = nπ \pm \frac{π}{4}$
Also, $2 \leq s in x + 3 \leq 4$
$\Rightarrow s in x + 3 = π - \frac{π}{4}, π + \frac{π}{4}$
$\Rightarrow s in x = \frac{3 π}{4} - 3$ or $\frac{5 π}{4} - 3$

Q. The number of solutions of the equation cot2sin⁡x+3=1 in 0,3π is equal to

2

4

6

8

cot2sin⁡x+3=1=cot2⁡4π​ ⇒sinx+3=nπ±4π​ Also, 2≤sinx+3≤4 ⇒sinx+3=π−4π​,π+4π​ ⇒sinx=43π​−3 or 45π​−3

Q. The number of solutions of the equation $co t^{2} s in x + 3 = 1$ in $0, 3 π$ is equal to

$2$

$4$

$6$

$8$

$co t^{2} s in x + 3 = 1 = co t^{2} \frac{π}{4}$
$\Rightarrow s in x + 3 = nπ \pm \frac{π}{4}$
Also, $2 \leq s in x + 3 \leq 4$
$\Rightarrow s in x + 3 = π - \frac{π}{4}, π + \frac{π}{4}$
$\Rightarrow s in x = \frac{3 π}{4} - 3$ or $\frac{5 π}{4} - 3$