Let S = {x ∈ (-π, π) : x ≠ 0, ±π/2}. The sum of all distinct solutions of the equation √3 sec x + cosec x + 2(tan x - cot x) = 0 in the set S is equal to
S = { x ∈ ( − π , π ) : x ≠ 0 , ± π 2 } S = \{x \in (-\pi, \pi) : x \neq 0,\ \pm\frac{\pi}{2}\} S = { x ∈ ( − π , π ) : x = 0 , ± 2 π }
3 sec x + cosec x + 2 ( tan x − cot x ) = 0 \sqrt{3}\,\sec x + \operatorname{cosec} x + 2(\tan x - \cot x) = 0 3 sec x + cosec x + 2 ( tan x − cot x ) = 0
To find: ∑ ( all distinct solutions of the equation that lie in S ) \text{To find: } \sum (\text{all distinct solutions of the equation that lie in } S) To find: ∑ ( all distinct solutions of the equation that lie in S )
sec x = 1 cos x , cosec x = 1 sin x \sec x = \frac{1}{\cos x}, \quad \operatorname{cosec} x = \frac{1}{\sin x} sec x = c o s x 1 , cosec x = s i n x 1
tan x = sin x cos x , cot x = cos x sin x \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} tan x = c o s x s i n x , cot x = s i n x c o s x
3 sec x + cosec x + 2 ( tan x − cot x ) = 0 \sqrt{3}\,\sec x + \operatorname{cosec} x + 2(\tan x - \cot x) = 0 3 sec x + cosec x + 2 ( tan x − cot x ) = 0
× sin x cos x ( ≠ 0 on S ) \times\, \sin x \cos x \quad (\neq 0 \text{ on } S) × sin x cos x ( = 0 on S )
3 sin x + cos x + 2 ( sin 2 x − cos 2 x ) = 0 \sqrt{3}\sin x + \cos x + 2(\sin^2 x - \cos^2 x) = 0 3 sin x + cos x + 2 ( sin 2 x − cos 2 x ) = 0
sin 2 x − cos 2 x = − cos 2 x \sin^2 x - \cos^2 x = -\cos 2x sin 2 x − cos 2 x = − cos 2 x
3 sin x + cos x − 2 cos 2 x = 0 \sqrt{3}\sin x + \cos x - 2\cos 2x = 0 3 sin x + cos x − 2 cos 2 x = 0
2 cos 2 x = 3 sin x + cos x 2\cos 2x = \sqrt{3}\sin x + \cos x 2 cos 2 x = 3 sin x + cos x
3 sin x + cos x = 2 sin ( x + π 6 ) \sqrt{3}\sin x + \cos x = 2\sin\!\left(x + \frac{\pi}{6}\right) 3 sin x + cos x = 2 sin ( x + 6 π )
2 cos 2 x = 2 sin ( x + π 6 ) 2\cos 2x = 2\sin\!\left(x + \frac{\pi}{6}\right) 2 cos 2 x = 2 sin ( x + 6 π )
cos 2 x = sin ( x + π 6 ) \cos 2x = \sin\!\left(x + \frac{\pi}{6}\right) cos 2 x = sin ( x + 6 π )
sin ( x + π 6 ) = cos ( π 2 − x − π 6 ) \sin\!\left(x+\frac{\pi}{6}\right) = \cos\!\left(\frac{\pi}{2} - x - \frac{\pi}{6}\right) sin ( x + 6 π ) = cos ( 2 π − x − 6 π )
= cos ( π 3 − x ) = cos ( x − π 3 ) = \cos\!\left(\frac{\pi}{3} - x\right) = \cos\!\left(x - \frac{\pi}{3}\right) = cos ( 3 π − x ) = cos ( x − 3 π )
cos 2 x = cos ( x − π 3 ) \cos 2x = \cos\!\left(x - \frac{\pi}{3}\right) cos 2 x = cos ( x − 3 π )
cos A = cos B ⇒ A = 2 n π ± B \cos A = \cos B \Rightarrow A = 2n\pi \pm B cos A = cos B ⇒ A = 2 nπ ± B
2 x = 2 n π ± ( x − π 3 ) 2x = 2n\pi \pm \left(x - \frac{\pi}{3}\right) 2 x = 2 nπ ± ( x − 3 π )
2 x = 2 n π + x − π 3 2x = 2n\pi + x - \frac{\pi}{3} 2 x = 2 nπ + x − 3 π
x = 2 n π − π 3 x = 2n\pi - \frac{\pi}{3} x = 2 nπ − 3 π
2 x = 2 n π − x + π 3 2x = 2n\pi - x + \frac{\pi}{3} 2 x = 2 nπ − x + 3 π
3 x = 2 n π + π 3 3x = 2n\pi + \frac{\pi}{3} 3 x = 2 nπ + 3 π
x = 2 n π 3 + π 9 x = \frac{2n\pi}{3} + \frac{\pi}{9} x = 3 2 nπ + 9 π
n = 0 : x = − π 3 ∈ ( − π , π ) ✓ n=0:\ x = -\frac{\pi}{3} \in (-\pi, \pi)\ \checkmark n = 0 : x = − 3 π ∈ ( − π , π ) ✓
n = ± 1 , ± 2 , … : ∣ x ∣ > π (reject) n=\pm 1,\pm 2,\ldots:\ |x|>\pi \ \text{(reject)} n = ± 1 , ± 2 , … : ∣ x ∣ > π (reject)
n = 0 : x = π 9 n=0:\ x = \frac{\pi}{9} n = 0 : x = 9 π
n = 1 : x = 2 π 3 + π 9 = 7 π 9 n=1:\ x = \frac{2\pi}{3} + \frac{\pi}{9} = \frac{7\pi}{9} n = 1 : x = 3 2 π + 9 π = 9 7 π
n = − 1 : x = − 5 π 9 n=-1:\ x = -\frac{5\pi}{9} n = − 1 : x = − 9 5 π
n = 2 : x = 13 π 9 > π (reject) n=2:\ x = \frac{13\pi}{9} > \pi \ \text{(reject)} n = 2 : x = 9 13 π > π (reject)
x ∈ { − π 3 , π 9 , 7 π 9 , − 5 π 9 } x \in \left\{ -\frac{\pi}{3},\ \frac{\pi}{9},\ \frac{7\pi}{9},\ -\frac{5\pi}{9} \right\} x ∈ { − 3 π , 9 π , 9 7 π , − 9 5 π }
None equals 0 or ± π 2 ✓ \text{None equals } 0 \text{ or } \pm\frac{\pi}{2}\ \checkmark None equals 0 or ± 2 π ✓
Sum = − π 3 + π 9 + 7 π 9 − 5 π 9 \text{Sum} = -\frac{\pi}{3} + \frac{\pi}{9} + \frac{7\pi}{9} - \frac{5\pi}{9} Sum = − 3 π + 9 π + 9 7 π − 9 5 π
= − 3 π + π + 7 π − 5 π 9 = \frac{-3\pi + \pi + 7\pi - 5\pi}{9} = 9 − 3 π + π + 7 π − 5 π
= 0 9 = 0 = \frac{0}{9} = 0 = 9 0 = 0
Sum = 0 ⇒ option (3) \boxed{\text{Sum} = 0} \quad \Rightarrow \text{option (3)} Sum = 0 ⇒ option (3)
