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When we try to find out the roots in an equation, especially in case of a trigonometric equation, there can be several cases. it is extremely important to be careful while carrying out various algebraic operations because there are instances when we can cancel certain roots while sometimes we cannot.
Root loss in simple words means that some roots might be missed while performing some algebraic operations. Hence, it is very essential to be careful while carrying out algebraic operations. This usually occurs during cancellation of similar terms on both the sides. While performing the cancellation operation, we must note that we don’t miss out any root. To illustrate this, we consider the following example:
sin θ[A] = sin θ[B] where A and B are trigonometric functions.
So if we cancel sin θ then we lose the solutions, sin θ = 0
i.e. θ = n π, where n = 0, ±1, ±2, ………
Hence, particularly in case of trigonometric functions, the cancellation cannot be carried out directly.
The literal meaning of extraneous is extra. When we perform the operation of squaring, we may get some extra roots. So, especially in case of trigonometric functions, squaring should be avoided as far as possible and even if it is inevitable, one must check for the extra roots.
Illustration: sec θ – 1 = (√2 - 1) tan θ
Solution: We know the relation sec^{2} θ = tan^{2} θ + 1. So we can utilize this relation by squaring both the sides of the equation.
⇒ sec θ = (√2-1) tan θ + 1
or sec^{2} θ = (√2-1)^{2} tan^{2} θ + 1 + 2 (√2-1) tan θ
or 1 + tan^{2} θ = (2+1-2√2) tan^{2} θ + 1 + 2 (√2-1) tan θ
or 2(1-√2) tan^{2} θ + 2(√2-1) tan θ = 0
or 2(1-√2) tan θ (tan θ – 1) = 0
either tan θ = 0 ⇒ θ = n π where n = 0, ±1, ±2 ………
or tan θ – 1 = 0 ⇒ θ = n π + π/4. Where n = 0, ±1, ±2, ……
Remark: It is important to note here that all the solutions obtained are not the solutions of the original equation but of its squared form. Since squaring yields many extra roots so it is essential to check out the actual solutions of the original equation.
(i) For θ = nπ
sec (nπ) – 1 = (√2-1) tan (nπ)
(± 1) – 1 = 0
True only when n is even. Hence for odd n, an extra solution has been obtained which does not satisfy the original equation. So, it should be excluded.
⇒ Only θ = 2kπ is a solution
(ii) For θ = n π + π/4
sec (n π + π/4) – 1 = (√2-1) tan (n π + π/4)
√2-1 = (√2-1).
Yes, it does satisfy the given equation. Hence it’s a solution
Hence the general solution is
θ = (2mπ), (n π + π/4), where n, m = 0, ±1, ±2, ……
Remark: Sometimes the question demands the solution in a particular region. In that case the roots which though are solutions of the original equation become extra if they don’t lie in the specified restricted region.
Illustration: The number of values of x in the interval [0, 5π] satisfying the equation 3 sin^{2}x – 7 sin x + 2 = 0, is
(1) 0 (2) 5
(3) 6 (4) 10
Solution: The given equation is 3sin^{2}x – 7 sin x + 2 = 0.
This can be written as 3sin^{2}x – 6 sin x - sin x + 2 = 0.
This gives 3sin x (sin x - 2) – (sin x - 2) = 0
Hence, (3sin x – 1) (sin x - 2) = 0.
This gives sin x = 1/3. (Since sin x = 2 is rejected)
Hence, x = nπ + (-1)n sin^{-1} 1/3, n ∈ I
For 0 ≤ n ≤ 5, x ∈ [0, 5π]
There are six values of x ∈ [0, 5π] which satisfy the equation 3sin2x – 7 sin x + 2 = 0.
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