{"id":9466,"date":"2022-01-14T16:14:47","date_gmt":"2022-01-14T10:44:47","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9466"},"modified":"2022-01-14T16:25:27","modified_gmt":"2022-01-14T10:55:27","slug":"general-solution-of-the-equation-of-form-a-costheta-b-sin-theta-c","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/general-solution-of-the-equation-of-form-a-costheta-b-sin-theta-c\/","title":{"rendered":"General Solution of the Equation of form \\(a cos\\theta\\) + \\(b sin \\theta\\) = c"},"content":{"rendered":"

General Solution of the equation \\(a cos\\theta\\) + \\(b sin \\theta\\) = c,<\/h2>\n

where a, b, c \\(\\in\\) R such that | c | \\(\\le\\) \\(\\sqrt{a^2 + b^2}\\).<\/p>\n

To solve this type of equations, we first reduce them in the form \\(cos \\theta\\) = \\(cos \\alpha\\),\u00a0 or\u00a0 \\(sin \\theta\\) = \\(sin \\alpha\\).<\/p>\n

The following algorithm provides the method of solution.<\/p>\n

Algorithm :<\/strong><\/p>\n

Step 1 : Obtain the equation \\(a cos\\theta\\) + \\(b sin \\theta\\) = c<\/p>\n

Step 2 : Put a = \\(r cos\\alpha\\)\u00a0 and\u00a0 b = \\(r sin \\alpha\\) ,\u00a0 where\u00a0 r = \\(\\sqrt{a^2 + b^2}\\) and \\(tan \\alpha\\) = \\(b\\over a\\)\u00a0 i.e. \\(\\alpha\\) = \\(tan^{-1} (b\/a)\\).<\/p>\n

Step 3 : Using the substitution in step 2, the equation reduces to \\(r cos(\\theta – \\alpha)\\) = c \\(\\implies\\) \\(cos(\\theta – \\alpha)\\) = \\(c\\over r\\) = \\(cos \\beta\\) (say).<\/p>\n

Step 4 : Solve the equation obtained in step 3 by using the general formula to solve trigonometric equations<\/a>.<\/p>\n

Example<\/strong><\/span> : Solve : \\(\\sqrt{3} cos \\theta\\) + \\(sin \\theta\\) = \\(\\sqrt{2}\\).<\/p>\n

Solution<\/strong> <\/span>: We have,<\/p>\n

\\(\\sqrt{3} cos \\theta\\) + \\(sin \\theta\\) = \\(\\sqrt{2}\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………….(i)<\/p>\n

This is of the form \\(a cos\\theta\\) + \\(b sin \\theta\\) = c, where a = \\(\\sqrt{3}\\), b = 1 and c = \\(\\sqrt{2}\\).<\/p>\n

Let a = \\(r cos\\alpha\\) and b = \\(r sin\\alpha\\). Then,<\/p>\n

\\(\\sqrt{3}\\) = \\(r cos\\alpha\\)\u00a0 and\u00a0 1 = \\(r sin\\alpha\\)<\/p>\n

\\(\\implies\\)\u00a0 r = \\(\\sqrt{a^2 + b^2}\\) = \\(\\sqrt{(\\sqrt{3})^2 + 1^2}\\) = 2 and \\(tan \\alpha\\) = \\(1\\over \\sqrt{3}\\) \\(\\implies\\)\u00a0 \\(\\alpha\\) = \\(\\pi\\over 6\\)<\/p>\n

Substituting a = \\(\\sqrt{3}\\) = \\(r cos\\alpha\\)\u00a0 and\u00a0 b = 1 = \\(r sin \\alpha\\) in the equation (i) it reduces to<\/p>\n

\\(r cos\\alpha cos\\theta\\) + \\(r sin\\alpha sin\\theta\\) = \\(\\sqrt{2}\\)<\/p>\n

\\(\\implies\\) \\(r cos(\\theta – \\alpha)\\) = \\(\\sqrt{2}\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(2 cos(\\theta – {\\pi\\over 6}\\) = \\(\\sqrt{2}\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(cos(\\theta – {\\pi\\over 6})\\) = \\(1\\over \\sqrt{2}\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(cos(\\theta – {\\pi\\over 6})\\)\u00a0 = \\(cos{\\pi\\over 4}\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(\\theta – {\\pi\\over 6}\\) = \\(2n\\pi \\pm {\\pi\\over 4}\\),\u00a0 n \\(\\in\\)\u00a0 Z.<\/p>\n

\\(\\implies\\)\u00a0 \\(\\theta\\) = \\(2n\\pi \\pm {\\pi\\over 4} + {\\pi\\over 6}\\),\u00a0 n \\(\\in\\)\u00a0 Z.<\/p>\n

\\(\\implies\\)\u00a0 \\(\\theta\\) = \\(2n\\pi + {\\pi\\over 4} + {\\pi\\over 6}\\)\u00a0 or,\u00a0 \\(\\theta\\) = \\(2n\\pi – {\\pi\\over 4} + {\\pi\\over 6}\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(\\theta\\) = \\(2n\\pi + {5\\pi\\over 12}\\)\u00a0 or\u00a0 \\(2n\\pi – {\\pi\\over 12}\\)<\/p>\n

Hence,\u00a0 \\(\\theta\\) = \\(2n\\pi + {5\\pi\\over 12}\\)\u00a0 or,\u00a0 \\(\\theta\\) = \\(2n\\pi – {5\\pi\\over 12}\\),\u00a0 where n \\(\\in\\)\u00a0 Z.<\/p>\n","protected":false},"excerpt":{"rendered":"

General Solution of the equation \\(a cos\\theta\\) + \\(b sin \\theta\\) = c, where a, b, c \\(\\in\\) R such that | c | \\(\\le\\) \\(\\sqrt{a^2 + b^2}\\). To solve this type of equations, we first reduce them in the form \\(cos \\theta\\) = \\(cos \\alpha\\),\u00a0 or\u00a0 \\(sin \\theta\\) = \\(sin \\alpha\\). The following algorithm …<\/p>\n

General Solution of the Equation of form \\(a cos\\theta\\) + \\(b sin \\theta\\) = c<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[29],"tags":[],"yoast_head":"\nGeneral Solution of the Equation of form \\(a cos\\theta\\) + \\(b sin \\theta\\) = c<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/general-solution-of-the-equation-of-form-a-costheta-b-sin-theta-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"General Solution of the Equation of form \\(a cos\\theta\\) + \\(b sin \\theta\\) = c\" \/>\n<meta property=\"og:description\" content=\"General Solution of the equation (a costheta) + (b sin theta) = c, where a, b, c (in) R such that | c | (le) (sqrt{a^2 + b^2}). 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