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) cos x + cos 2x + cos 3x = 0
Let us simplify,
cos x + cos 2x + cos 3x = 0
we shall rearrange and use transformation formula
cos 2x + (cos x + cos 3x) = 0
by using the formula, cos A + cos B = 2 cos (A+B)/2 cos(A-B)/2
cos 2x + 2 cos (3x+x)/2 cos (3x-x)/2 = 0
cos 2x + 2cos 2x cos x = 0
cos 2x ( 1 + 2 cos x) = 0
cos 2x = 0 or 1 + 2cos x = 0
cos 2x = cos 0 or cos x = -1/2
cos 2x = cos π/2 or cos x = cos (π – π/3)
cos 2x = cos π/2 or cos x = cos (2π/3)
2x = (2n + 1) π/2 or x = 2mπ ± 2π/3
x = (2n + 1) π/4 or x = 2mπ ± 2π/3
∴ the general solution is
x = (2n + 1) π/4 or 2mπ ± 2π/3, where m, n ϵ Z.
(ii) cos x + cos 3x – cos 2x = 0
Let us simplify,
cos x + cos 3x – cos 2x = 0
we shall rearrange and use transformation formula
cos x – cos 2x + cos 3x = 0
– cos 2x + (cos x + cos 3x) = 0
By using the formula, cos A + cos B = 2 cos (A+B)/2 cos(A-B)/2
– cos 2x + 2 cos (3x+x)/2 cos (3x-x)/2 = 0
– cos 2x + 2cos 2x cos x = 0
cos 2x ( -1 + 2 cos x) = 0
cos 2x = 0 or -1 + 2cos x = 0
cos 2x = cos 0 or cos x = 1/2
cos 2x = cos π/2 or cos x = cos (π/3)
2x = (2n + 1) π/2 or x = 2mπ ± π/3
x = (2n + 1) π/4 or x = 2mπ ± π/3
∴ the general solution is
x = (2n + 1) π/4 or 2mπ ± π/3, where m, n ϵ Z.
(iii) sin x + sin 5x = sin 3x
Let us simplify,
sin x + sin 5x = sin 3x
sin x + sin 5x – sin 3x = 0
we shall rearrange and use transformation formula
– sin 3x + sin x + sin 5x = 0
– sin 3x + (sin 5x + sin x) = 0
By using the formula, sin A + sin B = 2 sin (A+B)/2 cos(A-B)/2
– sin 3x + 2 sin (5x+x)/2 cos (5x-x)/2 = 0
2sin 3x cos 2x – sin 3x = 0
sin 3x ( 2cos 2x – 1) = 0
sin 3x = 0 or 2cos 2x – 1 = 0
sin 3x = sin 0 or cos 2x = 1/2
sin 3x = sin 0 or cos 2x = cos π/3
3x = nπ or 2x = 2mπ ± π/3
x = nπ/3 or x = mπ ± π/6
∴ the general solution is
x = nπ/3 or mπ ± π/6, where m, n ϵ Z.
(iv) cos x cos 2x cos 3x = 1/4
Let us simplify,
cos x cos 2x cos 3x = 1/4
4 cos x cos 2x cos 3x – 1 = 0
By using the formula,
2 cos A cos B = cos (A + B) + cos (A – B)
2(2cos x cos 3x) cos 2x – 1 = 0
2(cos 4x + cos 2x) cos2x – 1 = 0
2(2cos2 2x – 1 + cos 2x) cos 2x – 1 = 0[using cos 2A = 2cos2A – 1]
4cos3 2x – 2cos 2x + 2cos2 2x– 1 = 0
2cos2 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0
(2cos2 2x – 1) (2 cos 2x + 1) = 0
So,
2cos 2x + 1 = 0 or (2cos2 2x – 1) = 0
cos 2x = -1/2 or cos 4x = 0 [using cos 2θ = 2cos2θ– 1]
cos 2x = cos (π – π/3) or cos 4x = cos π/2
cos 2x = cos 2π/3 or cos 4x = cos π/2
2x = 2mπ ± 2π/3 or 4x = (2n + 1) π/2
x = mπ ± π/3 or x = (2n + 1) π/8
∴ the general solution is
x = mπ ± π/3 or (2n + 1) π/8, where m, n ϵ Z.
(v) cos x + sin x = cos 2x + sin 2x
Let us simplify,
cos x + sin x = cos 2x + sin 2x
upon rearranging we get,
cos x – cos 2x = sin 2x – sin x
By using the formula,
sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
cos A – cos B = – 2 sin (A+B)/2 sin (A-B)/2
So,
-2 sin (2x+x)/2 sin (2x-x)/2 = 2 cos (2x+x)/2 sin (2x-x)/2
2 sin 3x/2 sin x/2 = 2 cos 3x/2 sin x/2
Sin x/2 (sin 3x/2 – cos 3x/2) = 0
So,
Sin x/2 = 0 or sin 3x/2 = cos 3x/2
Sin x/2 = sin mπ or sin 3x/2 / cos 3x/2 = 0
Sin x/2 = sin mπ or tan 3x/2 = 1
Sin x/2 = sin mπ or tan 3x/2 = tan π/4
x/2 = mπ or 3x/2 = nπ + π/4
x = 2mπ or x = 2nπ/3 + π/6
∴ the general solution is
x = 2mπ or 2nπ/3 + π/6, where m, n ϵ Z.
(vi) sin x + sin 2x + sin 3x = 0
Let us simplify,
sin x + sin 2x + sin 3x = 0
we shall rearrange and use transformation formula
sin 2x + sin x + sin 3x = 0
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
So,
Sin 2x + 2 sin (3x+x)/2 cos (3x-x)/2 = 0
Sin 2x + 2sin 2x cos x = 0
Sin 2x (2 cos x + 1) = 0
Sin 2x = 0 or 2cos x + 1 = 0
Sin 2x = sin 0 or cos x = -1/2
Sin 2x = sin 0 or cos x = cos (π – π/3)
Sin 2x = sin 0 or cos x = cos 2π/3
2x = nπ or x = 2mπ ± 2π/3
x = nπ/2 or x = 2mπ ± 2π/3
∴ the general solution is
x = nπ/2 or 2mπ ± 2π/3, where m, n ϵ Z.
(vii) sin x + sin 2x + sin 3x + sin 4x = 0
Let us simplify,
sin x + sin 2x + sin 3x + sin 4x = 0
we shall rearrange and use transformation formula
sin x + sin 3x + sin 2x + sin 4x = 0
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
So,
2 sin (3x+x)/2 cos (3x-x)/2 + 2 sin (4x+2x)/2 cos (4x-2x)/2= 0
2 sin 2x cos x + 2 sin 3x cos x = 0
2cos x (sin 2x + sin 3x) = 0
Again by using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
we get,
2cos x (2 sin (3x+2x)/2 cos (3x-2x)/2) = 0
2cos x (2 sin 5x/2 cos x/2) = 0
4 cos x sin 5x/2 cos x/2 = 0
So,
Cos x = 0 or sin 5x/2 = 0 or cos x/2 = 0
Cos x = cos 0 or sin 5x/2 = sin 0 or cos x/2 = cos 0
Cos x = cos π/2 or sin 5x/2 = kπ or cos x/2 = cos (2p + 1)π/2
x = (2n + 1) π/2 or 5x/2 = kπ or x/2 = (2p + 1) π/2
x = (2n + 1) π/2 or x = 2kπ/5 or x = (2p + 1)
x = nπ + π/2 or x = 2kπ/5 or x = (2p + 1)
∴ the general solution is
x = nπ + π/2 or x = 2kπ/5 or x = (2p + 1), where n, k,p ϵ Z.
(viii) sin 3x – sin x = 4 cos2 x – 2
Let us simplify,
sin 3x – sin x = 4 cos2 x – 2
sin 3x – sin x = 2(2 cos2 x – 1)
sin 3x – sin x = 2 cos 2x [since, cos 2A = 2cos2 A– 1]
By using the formula,
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
So,
2 cos (3x+x)/2 sin (3x-x)/2 = 2 cos 2x
2 cos 2x sin x – 2 cos 2x = 0
2 cos 2x (sin x – 1) = 0
Then,
2 cos 2x = 0 or sin x – 1 = 0
Cos 2x = 0 or sin x = 1
Cos 2x = cos 0 or sin x = sin 1
Cos 2x = cos 0 or sin x = sin π/2
2x = (2n + 1) π/2 or x = mπ + (-1) m π/2
x = (2n + 1) π/4 or x = mπ + (-1) m π/2
∴ the general solution is
x = (2n + 1) π/4 or mπ + (-1) m π/2,where m, n ϵ Z.
(ix) sin 2x – sin 4x + sin 6x = 0
Let us simplify,
sin 2x – sin 4x + sin 6x = 0
we shall rearrange and use transformation formula
– sin 4x + sin 6x + sin 2x = 0
By using the formula,
sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
we get,
– sin 4x + 2 sin (6x+2x)/2 cos (6x-2x)/2 = 0
– sin 4x + 2 sin 4x cos 2x = 0
Sin 4x (2 cos 2x – 1) = 0
So,
Sin 4x = 0 or 2 cos 2x – 1 = 0
Sin 4x = sin 0 or cos 2x = 1/2
Sin 4x = sin 0 or cos 2x = π/3
4x = nπ or 2x = 2mπ ± π/3
x = nπ/4 or x = mπ ± π/6
∴ the general solution is
x = nπ/4 or mπ ± π/6, where m, n ϵ Z.