1
BITSAT 2023
+3
-1

If $$n$$ is the number of solutions of the equation $$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$$ and $$S$$ is the sum of all these solutions, then the ordered pair $$(n, S)$$ is

A
$$(3,13 \pi / 9)$$
B
$$(2,2 \pi / 3)$$
C
$$(2,8 \pi / 9)$$
D
$$(3,5 \pi / 3)$$
2
BITSAT 2022
+3
-1

The sum of all the solution of the equation $$\cos \theta \cos \left( {{\pi \over 3} + \theta } \right)\cos \left( {{\pi \over 3} - \theta } \right) = {1 \over 4},\theta \in [0,6\pi ]$$

A
15$$\pi$$
B
30$$\pi$$
C
$${{100\pi } \over 3}$$
D
None of these
3
BITSAT 2021
+3
-1

If $${\cos ^3}x\,.\,\sin 2x = \sum\limits_{m = 1}^n {{a_m}\sin mx}$$ is identity in x, then

A
$${a_3} = {3 \over 8},{a_2} = 0$$
B
$$n = 6,{a_1} = {1 \over 2}$$
C
$$n = 5,{a_1} = {3 \over 4}$$
D
$$\sum {{a_m} = {1 \over 4}}$$
4
BITSAT 2021
+3
-1

Total number of solutions of $$\left| {\cot x} \right| = \cot x + {1 \over {\sin x}},x \in [0,3\pi ]$$ is equal to

A
1
B
2
C
3
D
0
EXAM MAP
Medical
NEET