Trigonometric Equations Solving Strategies

Here, you will learn trigonometric equations solving strategies i.e by factorisation and reducing it to a quadratic equation etc.

Let’s begin –

Different Strategies for Trigonometric Equations Solving

(a) Solving trigonometric equations by factorisation

Example : If \(1\over 6\)sin x, cos x, tan x are in G.P. then the general solution for x is

Solution : Since \(1\over 6\)sin x, cos x, tan x are in G.P.

\(\implies\)   \(cos^2x\) = \(1\over 6\)sin x.tan x

\(\implies\)   6\(cos^3x\) + \(cos^2x\) – 1 = 0

\(\therefore\)   (2cos x – 1)(3\(cos^2x\) + 2cos x + 1) = 0

\(\implies\)   cos x = \(1\over 2\)   (other values of cos x are imaginary)

\(\implies\)   cos x = cos\(\pi\over 3\)   \(\implies\)   x = 2n\(\pi\) \(\pm\) \(\pi\over 3\)

(b) Solving trigonometric equations by reducing it to a quadratic equation

Example : Find the number of solutions of tan x + sec x = 2cos x in [0,\(2\pi\)].

Solution : Here, tan x + sec x = 2cos x   \(\implies\)   sin x + 1 = 2\(cos^2x\)

\(\implies\)   2\(sin^2x\) + sin x – 1 = 0   \(\implies\)   sin x = \(1\over 2\), -1

But sin x = -1   \(\implies\)   x = \(3\pi\over 2\)   for which tan x + sec x = 2cos x is not defined.

Thus sin x = \(1\over 2\)   \(\implies\)   x = \(\pi\over 6\), \(5\pi\over 6\)

\(\implies\)   number of solutions of tan x + sec x = 2cos x is 2.

(c) Solving trigonometric equations by introducing an auxilliary argument

Example : Find the number of distinct solutions of sec x + tan x = \(\sqrt{3}\), where 0 \(\le\) x \(\le\) \(3\pi\).

Solution : Here, sec x + tan x = \(\sqrt{3}\)   \(\implies\)   sin x + 1 = \(\sqrt{3}\)cos x or \(\sqrt{3}\)cos x – sin x = 1

Dividing both sides by \(\sqrt{a^2+b^2}\) i.e. 2, we get

\(\implies\)   \(\sqrt{3}\over 2\)cos x – \(1\over 2\)sin x = \(1\over 2\)

\(\implies\)   cos\(\pi\over 6\)cos x – sin\(\pi\over 6\)sin x = \(1\over 2\)   \(\implies\)   cos(x + \(\pi\over 6\)) = \(1\over 2\)

As 0 \(\le\) x \(\le\) 3\(\pi\)   \(\implies\)   \(\pi\over 6\) \(\le\) x + \(\pi\over 6\) \(\le\) \(3\pi + {\pi\over 6}\)

\(\implies\)   x + \(\pi\over 6\) = \(\pi\over 3\), \(5\pi\over 3\), \(7\pi\over 3\)   \(\implies\)   \(\pi\over 6\), \(3\pi\over 2\), \(13\pi\over 6\).

But at x = \(3\pi\over 2\), tan x and sec x is not defined.

\(\therefore\)   Total number of solutions are 2.

(d) Solving trigonometric equations by transforming sum of trigonometric functions into product

Example : Solve : cos x + cos 3x + cos 5x + cos 7x = 0

Solution : We have cos x + cos 7x + cos 3x + cos 5x = 0

\(\implies\)   2cos 4x cos 3x + 2cos 4x cos x = 0   \(\implies\)   cos 4x(cos 3x + cos x) = 0

\(\implies\)   cos 4x(2cos 2x cos x) = 0

\(\implies\)   Either cos x = 0   \(\implies\)   x = (\(2n_1 + 1\))\(\pi\over 2\), \(n_1\) \(\in\) I

or   cos 2x = 0   \(\implies\)   x = (\(2n_2 + 1\))\(\pi\over 4\), \(n_2\) \(\in\) I

or   cos 4x = 0   \(\implies\)   x = (\(2n_3 + 1\))\(\pi\over 8\), \(n_3\) \(\in\) I

(e) Solving trigonometric equations by transforming a product into sum

Example : Solve : cosx cos2x cos 3x = \(1\over 4\)

Solution : (2cosx cos3x)cos2x = \(1\over 2\)   \(\implies\)   (cos2x + cos4x)cos2x = \(1\over 2\)

\(\implies\)   \(1\over 2\)[\(2cos^22x\) + 2cos4x cos2x] = \(1\over 2\)   \(\implies\)   1 + cos4x + 2cos4xcos2x = 1

\(\therefore\)   cos 4x(1 + 2cos2x) = 0

\(\implies\)   cos 4x = 0   or   (1 + 2cos2x) = 0

Now from the first equation : cos 4x = 0 = cos(\(\pi\over 2\))

\(\therefore\)   4x = (2n + 1)\(\pi\over 2\)   \(\implies\)   x = (2n + 1)\(\pi\over 8\), n \(\in\) I

for   n = 0, x = \(\pi\over 8\); n = 1, x = \(3\pi\over 8\); n = 2, x = \(5\pi\over 8\); n = 3, x = \(7\pi\over 8\);

and from the second equation : cos 2x = -\(1\over 2\) = -cos(\(\pi\over 3\)) = cos(\(\pi-{\pi\over 3}\)) = cos(\(2\pi\over 3\))

\(\therefore\)   2x = 2k\(\pi\) \(\pm\) \(2\pi\over 3\)   \(\therefore\)   x = k\(\pi\) \(\pm\) \(\pi\over 3\), k \(\in\) I

again for k = 0, x = \(\pi\over 3\); k = 1, x = \(2\pi\over 3\)   (\(\because\) 0 \(\le\) x \(\le\) \(\pi\))

\(\therefore\)   x = \(\pi\over 8\), \(\pi\over 3\), \(3\pi\over 8\), \(5\pi\over 8\), \(2\pi\over 3\), \(7\pi\over 8\)

Trigonometric Inequalities

There is no general rule to solve trigonometric inequalities and the same rules of algebra are valid provided the domain and range of trigonometric function should be kept in mind.

Example : Find the solution set of inequality sin x > 1/2.

Solution : when sin x = \(1\over 2\), the two values of x between 0 and \(2\pi\) are \(pi\over 6\) and \(5pi\over 6\).

From the graph of y = sinx, it is obvious that it is between 0 and \(2\pi\),

sinx > \(1\over 2\) for \(\pi/6\) < x < \(5\pi/6\)

Hence, sinx > 1/2

\(\implies\) \(2n\pi\) + \(\pi\over 6\) < x < \(2n\pi\) + \(5\pi\over 6\), n \(\in\) I

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