Question -
Answer -
The general solution of any trigonometric equation is givenas:
sin x = sin y, implies x = nπ + (– 1)n y,where n ∈ Z.
cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.
tan x = tan y, implies x = nπ + y, where n ∈ Z.
(i) sin 2x = √3/2
Let us simplify,
sin 2x = √3/2
= sin (π/3)
∴ the general solution is
2x = nπ + (-1)n π/3, where n ϵ Z.
x = nπ/2 + (-1)n π/6, where n ϵ Z.
(ii) cos 3x = 1/2
Let us simplify,
cos 3x = 1/2
= cos (π/3)
∴ the general solution is
3x = 2nπ ± π/3, where n ϵ Z.
x = 2nπ/3 ± π/9, where n ϵ Z.
(iii) sin 9x = sin x
Let us simplify,
Sin 9x – sin x = 0
Using transformation formula,
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
So,
= 2 cos (9x+x)/2 sin (9x-x)/2
=> cos 5x sin 4x = 0
Cos 5x = 0 or sin 4x = 0
Let us verify both the expressions,
Cos 5x = 0
Cos 5x = cos π/2
5x = (2n + 1)π/2
x = (2n + 1)π/10, where n ϵ Z.
sin 4x = 0
sin 4x = sin 0
4x = nπ
x = nπ/4, where n ϵ Z.
∴ the general solution is
x = (2n + 1)π/10 or nπ/4, where n ϵ Z.
(iv) sin 2x = cos 3x
Let us simplify,
sin 2x = cos 3x
cos (π/2 – 2x) = cos 3x [since, sin A = cos (π/2 – A)]
π/2 – 2x = 2nπ ± 3x
π/2 – 2x = 2nπ + 3x [or] π/2 – 2x = 2nπ – 3x
5x = π/2 + 2nπ [or] x = 2nπ – π/2
5x = π/2 (1 + 4n) [or] x = π/2 (4n – 1)
x = π/10 (1 + 4n) [or] x = π/2 (4n – 1)
∴ the general solution is
x = π/10 (4n + 1) [or] x = π/2 (4n – 1), wheren ϵ Z.
(v) tan x + cot 2x = 0
Let us simplify,
tan x = – cot 2x
tan x = – tan (π/2 – 2x) [since, cot A = tan (π/2 – A)]
tan x = tan (2x – π/2) [since, – tan A = tan -A]
x = nπ + 2x – π/2
x = nπ – π/2
∴ the general solution is
x = nπ – π/2, where n ϵ Z.
(vi) tan 3x = cot x
Let us simplify,
tan 3x = cot x
tan 3x = tan (π/2 – x) [since, cot A = tan (π/2 – A)]
3x = nπ + π/2 – x
4x = nπ + π/2
x = nπ/4 + π/8
∴ the general solution is
x = nπ/4 + π/8, where n ϵ Z.
(vii) tan 2x tan x = 1
Let us simplify,
tan 2x tan x = 1
tan 2x = 1/tan x
= cot x
tan 2x = tan (π/2 – x) [since, cot A = tan (π/2 – A)]
2x = nπ + π/2 – x
3x = nπ + π/2
x = nπ/3 + π/6
∴ the general solution is
x = nπ/3 + π/6, where n ϵ Z.
(viii) tan mx + cot nx = 0
Let us simplify,
tan mx + cot nx = 0
tan mx = – cot nx
= – tan (π/2 – nx) [since, cot A = tan (π/2 – A)]
tan mx = tan (nx + π/2) [since, – tan A = tan -A]
mx = kπ + nx + π/2
(m – n) x = kπ + π/2
(m – n) x = π (2k + 1)/2
x = π (2k + 1)/2(m – n)
∴ the general solution is
x = π (2k + 1)/2(m – n), where m, n, k ϵ Z.
(ix) tan px = cot qx
Let us simplify,
tan px = cot qx
tan px = tan (π/2 – qx) [since, cot A = tan (π/2 – A)]
px = nπ ± (π/2 – qx)
(p + q) x = nπ + π/2
x = nπ/(p+q) + π/2(p+q)
= π (2n +1)/ 2(p+q)
∴ the general solution is
x = π (2n +1)/ 2(p+q), where n ϵ Z.
(x) sin 2x + cos x = 0
Let us simplify,
sin 2x + cos x = 0
cos x = – sin 2x
cos x = – cos (π/2 – 2x) [since, sin A = cos (π/2 – A)]
= cos (π – (π/2 – 2x)) [since, -cos A = cos (π – A)]
= cos (π/2 + 2x)
x = 2nπ ± (π/2 + 2x)
So,
x = 2nπ + (π/2 + 2x) [or] x = 2nπ – (π/2 + 2x)
x = – π/2 – 2nπ [or] 3x = 2nπ – π/2
x = – π/2 (1 + 4n) [or] x = π/6 (4n – 1)
∴ the general solution is
x = – π/2 (1 + 4n), where n ϵ Z. [or] x = π/6 (4n– 1)
x = π/2 (4n – 1), where n ϵ Z. [or] x = π/6 (4n –1), where n ϵ Z.
(xi) sin x = tan x
Let us simplify,
sin x = tan x
sin x = sin x/cos x
sin x cos x = sin x
sin x (cos x – 1) = 0
So,
Sin x = 0 or cos x – 1 = 0
Sin x = sin 0 [or] cos x = 1
Sin x = sin 0 [or] cos x = cos 0
x = nπ [or] x = 2mπ
∴ the general solution is
x = nπ [or] 2mπ, where n, m ϵ Z.
(xii) sin 3x + cos 2x = 0
Let us simplify,
sin 3x + cos 2x = 0
cos 2x = – sin 3x
cos 2x = – cos (π/2 – 3x) [since, sin A = cos (π/2 – A)]
cos 2x = cos (π – (π/2 – 3x)) [since, -cos A = cos (π – A)]
cos 2x = cos (π/2 + 3x)
2x = 2nπ ± (π/2 + 3x)
So,
2x = 2nπ + (π/2 + 3x) [or] 2x = 2nπ – (π/2 + 3x)
x = -π/2 – 2nπ [or] 5x = 2nπ – π/2
x = -π/2 (1 + 4n) [or] x = π/10 (4n – 1)
x = – π/2 (4n + 1) [or] π/10 (4n – 1)
∴ the general solution is
x = – π/2 (4n + 1) [or] π/10 (4n – 1)
x = π/2 (4n – 1) [or] π/10 (4n – 1), where n ϵ Z.