We will learn how to solve trigonometric equation using formula.
Here we will use the following formulas to get the solution of the trigonometric equations.
(a) If sin θ = 0 then θ = nπ, where n = 0, ± 1, ± 2, ± 3, …….
(b) If cos θ = 0 then θ = (2n + 1) \(\frac{π}{2}\), where n = 0, ± 1, ± 2, ± 3, …….
(c) If cos θ = cos ∝ then θ = 2nπ ± ∝, where n = 0, ± 1, ± 2, ± 3, …….
(d) If sin θ = sin ∝ then θ = n π + (-1) \(^{n}\) ∝, where n = 0, ± 1, ± 2, ± 3, …….
(e) If a cos θ + b sin θ = c then θ = 2nπ + ∝ ± β, where cos β = \(\frac{c}{\sqrt{a^{2} + b^{2}}}\), cos ∝ = \(\frac{a}{\sqrt{a^{2} + b^{2}}}\) and sin ∝ = \(\frac{b}{\sqrt{a^{2} + b^{2}}}\), where n = 0, ± 1, ± 2, ± 3, …….
1. Solve tan x + sec x = √3. Also find values of x between 0° and 360°.
Solution:
tan x + sec x = √3
⇒ \(\frac{sin x}{cos x}\) + \(\frac{1}{cos x}\) = √3, where cos x ≠ 0
⇒ sin x + 1 = √3 cos x
⇒ √3 cos x - sin x = 1,
This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = -1 and c = 1.
⇒ Now dividing both sides by \(\sqrt{(\sqrt{3})^{2} + (1)^{2}}\)
⇒ \(\frac{√3}{2}\) cos x - \(\frac{1}{2}\)sin x = \(\frac{1}{2}\)
⇒ cos x cos \(\frac{π}{4}\) – sin x sin \(\frac{π}{6}\) = cos \(\frac{π}{3}\)
⇒ cos (x + \(\frac{π}{6}\)) = cos \(\frac{π}{3}\)
⇒ x + \(\frac{π}{6}\) = 2nπ ± \(\frac{π}{3}\), where n = 0, ± 1, ± 2, ± 3, …….
⇒ x = 2nπ ± \(\frac{π}{3}\) - \(\frac{π}{6}\), where n = 0, ± 1, ± 2, ± 3, …….
When we take minus sign with \(\frac{π}{3}\), we get
x = 2nπ - \(\frac{π}{3}\) - \(\frac{π}{6}\)
⇒ x = 2nπ - \(\frac{π}{2}\), so that cos x = cos (2nπ - \(\frac{π}{2}\)) = cos \(\frac{π}{2}\) = 0, which spoils the assumption cos x ≠ 0 (otherwise the given equation would be meaningless).
So, x = 2nπ + \(\frac{π}{3}\) - \(\frac{π}{6}\), where n = 0, ± 1, ± 2, ± 3, …….
⇒ x = 2nπ + \(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, ……. is the general
solution of the given equation tan x + sec x = √3.
The only solution between 0° and 360° is x = \(\frac{π}{6}\) = 30°
2. Find the general solutions of θ which satisfy the equation sec θ = - √2
Solution:
sec θ = - √2
⇒ cos θ = - \(\frac{1}{√2}\)
⇒ cos θ = - cos \(\frac{π}{4}\)
⇒ cos θ = cos (π - \(\frac{π}{4}\))
⇒ cos θ = cos \(\frac{3π}{4}\)
⇒ θ = 2nπ ± \(\frac{3π}{4}\), where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solutions of θ which satisfy the equation sec θ = - √2 is θ = 2nπ ± \(\frac{3π}{4}\), where, n = 0, ± 1, ± 2, ± 3, …….
3. Solve the equation 2 cos\(^{2}\) x + 3 sin x = 0
Solution:
2 cos\(^{2}\) x + 3 sin x = 0
⇒ 2(1 - sin\(^{2}\) x) + 3 sin x = 0
⇒ 2 – 2 sin\(^{2}\) x + 3 sin x = 0
⇒ 2 sin\(^{2}\) x – 3 sin x – 2 = 0
⇒ 2 sin\(^{2}\) x - 4 sin x + sin x – 2 = 0
⇒ 2 sin x(sin x - 2) + 1(sin – 2) = 0
⇒ (sin x - 2)(2 sin x + 1) = 0
⇒ Either sin x - 2 =0 or 2 sin x + 1 = 0
But sin x – 2 = 0 i.e., sin x = 2, which is not possible.
Now form 2 sin x + 1 = 0 we get
⇒ sin x = -½
⇒ sin x =- sin \(\frac{π}{6}\)
⇒ sin x = sin (π + \(\frac{π}{6}\))
⇒ sin x = sin \(\frac{7π}{6}\)
⇒ x = nπ + (1)\(^{n}\)\(\frac{7π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the solution for the equation 2 cos\(^{2}\) x + 3 sin x = 0 is x = nπ + (1)\(^{n}\)\(\frac{7π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….
Note: In the above trig equation we observe that there is more than one trigonometric function. So, the identities (sin \(^{2}\) θ + cos \(^{2}\) θ = 1) are required to reduce the given equation to a single function.
4. Find the general solutions of cos x + sin x = cos 2x + sin 2x
Solution:
cos x + sin x = cos 2x + sin 2x
⇒cos x - cos 2x - sin 2x + sin x = 0
⇒ (cos x - cos 2x) - (sin 2x - sin x) = 0
⇒ 2 sin \(\frac{3x}{2}\) sin \(\frac{x}{2}\) - 2 cos \(\frac{3x}{2}\) sin \(\frac{x}{2}\) = 0
⇒ sin \(\frac{x}{2}\) (sin \(\frac{3x}{2}\) - cos \(\frac{3x}{2}\)) = 0
Therefore, either, sin \(\frac{x}{2}\) = 0
⇒ \(\frac{x}{2}\)= nπ
⇒ x = 2nπ
or, sin \(\frac{3x}{2}\) - cos \(\frac{3x}{2}\) = 0
⇒ sin \(\frac{3x}{2}\) = cos \(\frac{3x}{2}\)
⇒ tan \(\frac{3x}{2}\) = 1
⇒ tan \(\frac{3x}{2}\) = tan \(\frac{π}{4}\)
⇒ \(\frac{3x}{2}\)= nπ + \(\frac{π}{4}\)
⇒ x = \(\frac{1}{3}\) (2nπ + \(\frac{π}{2}\)) = (4n + 1)\(\frac{π}{6}\)
Therefore, the general solutions of cos x + sin x = cos 2x + sin 2x are x = 2nπ and x = (4n+1)\(\frac{π}{6}\), Where, n = 0, ±1, ±2, …………………..
5. Find the general solutions of sin 4x cos 2x = cos 5x sin x
Solution:
sin 4x cos 2x = cos 5x sin x
⇒ 2 sin 4x cos 2x = 2 cos 5x sin x
⇒ sin 6x + sin 2x = sin 6x - sin 4x
⇒ sin 2x + sin 4x =0
⇒ 2sin 3x cos x =0
Therefore, either, sin 3x = 0 or, cos x = 0
i.e., 3x = nπ or, x = (2n + 1)\(\frac{π}{6}\)
⇒ x = \(\frac{nπ}{3}\) or, x = (2n + 1)\(\frac{π}{6}\)
Therefore, the general solutions of sin 4x cos 2x = cos 5x sin x are \(\frac{nπ}{3}\) and x = (2n + 1)\(\frac{π}{6}\)
11 and 12 Grade Math
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