{"id":3927,"date":"2021-08-10T11:04:06","date_gmt":"2021-08-10T11:04:06","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3927"},"modified":"2021-09-11T20:57:32","modified_gmt":"2021-09-11T15:27:32","slug":"trigonometric-equations-solving-strategies","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/trigonometric-equations-solving-strategies\/","title":{"rendered":"Trigonometric Equations Solving Strategies"},"content":{"rendered":"
Here, you will learn trigonometric equations solving strategies i.e by factorisation and reducing it to a quadratic equation etc.<\/p>\n
Let’s begin –<\/p>\n
Example : <\/span> If \\(1\\over 6\\)sin x, cos x, tan x are in G.P. then the general solution for x is\n <\/p>\n Solution : <\/span>Since \\(1\\over 6\\)sin x, cos x, tan x are in G.P. Example : <\/span> Find the number of solutions of tan x + sec x = 2cos x in [0,\\(2\\pi\\)].\n <\/p>\n Solution : <\/span>Here, tan x + sec x = 2cos x \\(\\implies\\) sin x + 1 = 2\\(cos^2x\\) Example : <\/span> Find the number of distinct solutions of sec x + tan x = \\(\\sqrt{3}\\), where 0 \\(\\le\\) x \\(\\le\\) \\(3\\pi\\). \n <\/p>\n Solution : <\/span>Here, sec x + tan x = \\(\\sqrt{3}\\) \\(\\implies\\) sin x + 1 = \\(\\sqrt{3}\\)cos x or \\(\\sqrt{3}\\)cos x – sin x = 1 Example : <\/span> Solve : cos x + cos 3x + cos 5x + cos 7x = 0\n <\/p>\n Solution : <\/span>We have cos x + cos 7x + cos 3x + cos 5x = 0 Example : <\/span> Solve : cosx cos2x cos 3x = \\(1\\over 4\\)\n <\/p>\n Solution : <\/span>(2cosx cos3x)cos2x = \\(1\\over 2\\) \\(\\implies\\) (cos2x + cos4x)cos2x = \\(1\\over 2\\) 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.<\/p>\n\n\n Example : <\/span> Find the solution set of inequality sin x > 1\/2.\n <\/p>\n Solution : <\/span>when sin x = \\(1\\over 2\\), the two values of x between 0 and \\(2\\pi\\) are \\(pi\\over 6\\) and \\(5pi\\over 6\\).
\n \\(\\implies\\) \\(cos^2x\\) = \\(1\\over 6\\)sin x.tan x
\n \\(\\implies\\) 6\\(cos^3x\\) + \\(cos^2x\\) – 1 = 0
\n \\(\\therefore\\) (2cos x – 1)(3\\(cos^2x\\) + 2cos x + 1) = 0
\n \\(\\implies\\) cos x = \\(1\\over 2\\) (other values of cos x are imaginary)
\n \\(\\implies\\) cos x = cos\\(\\pi\\over 3\\) \\(\\implies\\) x = 2n\\(\\pi\\) \\(\\pm\\) \\(\\pi\\over 3\\)
<\/p>\n\n\n(b) Solving trigonometric equations by reducing it to a quadratic equation<\/strong><\/h4>\n\n\n
\n \\(\\implies\\) 2\\(sin^2x\\) + sin x – 1 = 0 \\(\\implies\\) sin x = \\(1\\over 2\\), -1
\n But sin x = -1 \\(\\implies\\) x = \\(3\\pi\\over 2\\) for which tan x + sec x = 2cos x is not defined.
\n Thus sin x = \\(1\\over 2\\) \\(\\implies\\) x = \\(\\pi\\over 6\\), \\(5\\pi\\over 6\\)
\n \\(\\implies\\) number of solutions of tan x + sec x = 2cos x is 2.
<\/p>\n\n\n(c) Solving trigonometric equations by introducing an auxilliary argument<\/strong><\/h4>\n\n\n
\n Dividing both sides by \\(\\sqrt{a^2+b^2}\\) i.e. 2, we get
\n \\(\\implies\\) \\(\\sqrt{3}\\over 2\\)cos x – \\(1\\over 2\\)sin x = \\(1\\over 2\\)
\n \\(\\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\\)
\n As 0 \\(\\le\\) x \\(\\le\\) 3\\(\\pi\\) \\(\\implies\\) \\(\\pi\\over 6\\) \\(\\le\\) x + \\(\\pi\\over 6\\) \\(\\le\\) \\(3\\pi + {\\pi\\over 6}\\)
\n \\(\\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\\).
\n But at x = \\(3\\pi\\over 2\\), tan x and sec x is not defined.
\n \\(\\therefore\\) Total number of solutions are 2.
<\/p>\n\n\n(d) Solving trigonometric equations by transforming sum of trigonometric functions into product<\/strong><\/h4>\n\n\n
\n \\(\\implies\\) 2cos 4x cos 3x + 2cos 4x cos x = 0 \\(\\implies\\) cos 4x(cos 3x + cos x) = 0
\n \\(\\implies\\) cos 4x(2cos 2x cos x) = 0
\n \\(\\implies\\) Either cos x = 0 \\(\\implies\\) x = (\\(2n_1 + 1\\))\\(\\pi\\over 2\\), \\(n_1\\) \\(\\in\\) I
\n or cos 2x = 0 \\(\\implies\\) x = (\\(2n_2 + 1\\))\\(\\pi\\over 4\\), \\(n_2\\) \\(\\in\\) I
\n or cos 4x = 0 \\(\\implies\\) x = (\\(2n_3 + 1\\))\\(\\pi\\over 8\\), \\(n_3\\) \\(\\in\\) I
<\/p>\n\n\n(e) Solving trigonometric equations by transforming a product into sum<\/strong><\/h4>\n\n\n
\n \\(\\implies\\) \\(1\\over 2\\)[\\(2cos^22x\\) + 2cos4x cos2x] = \\(1\\over 2\\) \\(\\implies\\) 1 + cos4x + 2cos4xcos2x = 1
\n \\(\\therefore\\) cos 4x(1 + 2cos2x) = 0
\n \\(\\implies\\) cos 4x = 0 or (1 + 2cos2x) = 0
\n Now from the first equation : cos 4x = 0 = cos(\\(\\pi\\over 2\\))
\n \\(\\therefore\\) 4x = (2n + 1)\\(\\pi\\over 2\\) \\(\\implies\\) x = (2n + 1)\\(\\pi\\over 8\\), n \\(\\in\\) I
\n 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\\);
\n and from the second equation : cos 2x = -\\(1\\over 2\\) = -cos(\\(\\pi\\over 3\\)) = cos(\\(\\pi-{\\pi\\over 3}\\)) = cos(\\(2\\pi\\over 3\\))
\n \\(\\therefore\\) 2x = 2k\\(\\pi\\) \\(\\pm\\) \\(2\\pi\\over 3\\) \\(\\therefore\\) x = k\\(\\pi\\) \\(\\pm\\) \\(\\pi\\over 3\\), k \\(\\in\\) I
\n again for k = 0, x = \\(\\pi\\over 3\\); k = 1, x = \\(2\\pi\\over 3\\) (\\(\\because\\) 0 \\(\\le\\) x \\(\\le\\) \\(\\pi\\))
\n \\(\\therefore\\) x = \\(\\pi\\over 8\\), \\(\\pi\\over 3\\), \\(3\\pi\\over 8\\), \\(5\\pi\\over 8\\), \\(2\\pi\\over 3\\), \\(7\\pi\\over 8\\)
<\/p>\n\n\nTrigonometric Inequalities<\/h2>\n
\n From the graph of y = sinx, it is obvious that it is between 0 and \\(2\\pi\\),
\n sinx > \\(1\\over 2\\) for \\(\\pi\/6\\) < x < \\(5\\pi\/6\\)
\n Hence, sinx > 1\/2
\n \\(\\implies\\) \\(2n\\pi\\) + \\(\\pi\\over 6\\) < x < \\(2n\\pi\\) + \\(5\\pi\\over 6\\), n \\(\\in\\) I
<\/p>\n\n\n\n