# Trigonometric Equations and their General Solutions

## Trigonometric Equations and their General Solutions:

It has been discussed earlier that an infinite number of angles may have the same trigonometrical ratio. Thus, a trigonometrical equation has an infinite number of solutions. For example, if cos θ = √3/2, then since cos θ > 0, the angle θ lies either in the first quadrant or in the fourth quadrant.

Now, since cos θ = √3/2 = cos π/6, the least positive value of θ is π/6 radian or 30°.

Again, since cos θ = √3/2 = cos π/6 = cos (2π – π/6) = cos 11π/6, another value of θ satisfying the equation is 11π/6 radians. Similarly, as cos θ = cos π/6 = cos (4π – π/6) = cos 23π/6, another value of θ is obviously 23π/6. Thus, cos θ = √3/2 ⇒ θ = 2nπ + π/6, n ∈ z.

Also, cos θ = √3/2 = cos (- π/6) = cos (2nπ – π/6)

∴ cos θ = √3/2 ⇒ θ = 2nπ – π/6

Thus, for cos θ = √3/2, in general, we have

θ = 2nπ ± π/6, n ∈ z

### General solution for (i) sin θ = 0, (ii) cos θ = 0, (iii) tan θ = 0 and (iv) cot θ = 0:

(i) sin θ = 0

Since sin θ = 0 implies θ = 0°, 180°, 360°, ……… We have as general solution of the equation, θ = nπ, where n is any integer.

[In the above figure, sin θ = 0 ⇒ PN/OP = 0 ⇒ PN = 0]

(ii) cos θ = 0

Clearly, cos θ = 0 implies ON/OP = 0 (above figure)

⇒ ON = 0 which is possible only when θ = π/2, 3π/2, 5π/2, ……. i.e., when θ is an odd multiple of π/2.

Thus, cos θ = 0 ⇒ θ = (2n + 1)π/2, where n is any integer.

(iii) tan θ = 0

tan θ = 0 implies sin θ/cos θ = 0

⇒ sin θ = 0, which implies θ = nπ

Thus, the general solution for tan θ = 0 is θ = nπ, where n is any integer

(iv) cot θ = 0

cot θ = 0 implies cos θ/sin θ = 0

⇒ cos θ = 0, which gives θ = (2n + 1) π/2

Thus, as the general solution of the equation cot θ = 0, one gets θ = (2n + 1) π/2, where n is any integer.

### General solution of an equation of the form sin θ = c (-1 ≤ c ≤ 1):

We consider an angle α such that sin α = c and -π/2 ≤ α ≤ π/2. Then, sin θ = c = sin α implies sin θ – sin α = 0 ⇒ 2 cos (θ + α)/2 sin (θ – α)/2 = 0

Thus, either cos (θ + α)/2 = 0 or, sin (θ – α)/2 = 0

When cos (θ + α)/2 = 0, (θ + α)/2 = (2p + 1) π/2

⇒ θ = (2p + 1) π – α, where p ∈ z ……………..(i)

When, sin (θ – α)/2 = 0, (θ – α)/2 = rπ

⇒ θ = 2rπ + α, where r∈ z …………….(ii)

Now, (2p + 1) π implies an odd multiple of π whereas 2rπ means an even multiple of π. Thus, the general solution of sin θ = sin α (which is equal to c) is a solution that satisfies both (i) and (ii) which is evidently θ = (-1)n α + nπ, where n is any integer and -π/2 ≤ α ≤ π/2.

Note: (I) When c = 1, the equation becomes sin θ = 1 = sin π/2 = sin (2nπ + π/2)

∴ θ = 2nπ + π/2 = (4n + 1) π/2

Thus, when sin θ = 1, θ = (4n + 1) π/2

(II) When c = -1, the equation becomes sin θ = -1

⇒ sin θ = sin (-π/2) = sin (2nπ – π/2)

⇒ θ = 2nπ – π/2 = (4n – 1) π/2

Thus, when sin θ = -1, θ = (4n -1) π/2, n∈ z

(III) If we consider the equation cosec θ = c, |c| ≥ 1, then supposing α to be the least angle satisfying -π/2 ≤ α ≤ π/2, α ≠ 0, we have cosec α = c.

Thus, cosec θ = cosec α ⇒ sin θ = sin α

⇒ θ = (-1)n α + nπ, where n ∈ z

### General solution of cos θ = c (-1 ≤ c ≤ 1):

When 0 ≤ c ≤ 1, let cos α = c, where 0 ≤ α ≤ π/2. Also, when -1 ≤ c < 0, let cos α = c, where π/2 ≤ α ≤ π. Thus, in any case, cos θ = c = cos α

⇒ cos θ – cos α = 0 ⇒ 2 sin (θ + α)/2 sin (θ – α)/2 = 0

Thus, either sin (θ + α)/2 = 0 or, sin (θ – α)/2 = 0

When sin (θ + α)/2 = 0, (θ + α)/2 = pπ ⇒ θ = 2pπ – α, where p ∈ z ………(i)

When sin (θ – α)/2 = 0, (θ – α)/2 = rπ

⇒ θ = 2rπ + α, where r ∈ z ………….(ii)

Thus the general solution of cos θ = cos α is given by θ = 2nπ ± α, where n ∈ z.

Note: (I) When c = 1, the equation becomes cos θ = 1

⇒ cos θ = cos 0 = cos 2nπ ∴ θ =2nπ

Thus, when cos θ = 1, θ =2nπ, n ∈ z

(II) When c = -1, the equation reduces to cos θ = -1 ⇒ cos θ = cos π = cos (2n + 1) π, so that θ = (2n + 1) π.

Thus, when cos θ = -1, θ = (2n + 1) π, n ∈ z

(III) Let sec θ = c, |c| ≥ 1

Then, supposing α to be the least positive angle (0 ≤ α ≤ π) satisfying sec α = c, α ≠ π/2, we have, sec θ = sec α ⇒ cos θ = cos α

⇒ θ = 2nπ ± α, where n ∈ z

### General solution of tan θ = c (-∞ < c < ∞):

Let us consider α to be a definite positive or negative angle, with numerically the least measure for which tan α = c and -π/2 < α < π/2.

Then, tan θ = c = tan α gives tan θ – tan α = 0

⇒ sin θ/cos θ – sin α/cos α = 0 (-π/2 < α < π/2)

⇒ (sin θ cos α – cos θ sin α)/(cos θ cos α) = 0

⇒ sin (θ – α)/(cos θ cos α) = 0

⇒ sin (θ – α) = 0 (∵ cos θ cos α ≠ 0)

⇒ θ – α = nπ

⇒ θ = nπ + α, where n ∈ I

Thus, the general solution of the equation tan θ = tan α is θ = nπ + α, where n ∈ z and -π/2 < α < π/2.

Note: If the equation be cot θ = c (-∞ < c < ∞), then considering α to be the least angle satisfying cot α = c (-π/2 < α < π/2, α ≠ 0), then cot θ = cot α ⇒ tan θ = tan α and we get the same solution as above, namely θ = nπ + α, where n ∈ z.