How do you solve 2cos(3x + pi/4) =1?

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Here we will use the following formulas to get the solution of the trigonometric equations.

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(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, …….

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⇒ 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 =0Therefore, 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)\(\frac{π}{6}\)

 Trigonometric Equations

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