We have,<\/p>\n
\\(\\sqrt{3} cos \\theta\\) + \\(sin \\theta\\) = \\(\\sqrt{2}\\) ………….(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\\) and 1 = \\(r sin\\alpha\\)<\/p>\n
\\(\\implies\\) r = \\(\\sqrt{a^2 + b^2}\\) = \\(\\sqrt{(\\sqrt{3})^2 + 1^2}\\) = 2 and \\(tan \\alpha\\) = \\(1\\over \\sqrt{3}\\) \\(\\implies\\) \\(\\alpha\\) = \\(\\pi\\over 6\\)<\/p>\n
Substituting a = \\(\\sqrt{3}\\) = \\(r cos\\alpha\\) and 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\\) \\(2 cos(\\theta – {\\pi\\over 6}\\) = \\(\\sqrt{2}\\)<\/p>\n
\\(\\implies\\) \\(cos(\\theta – {\\pi\\over 6})\\) = \\(1\\over \\sqrt{2}\\)<\/p>\n
\\(\\implies\\) \\(cos(\\theta – {\\pi\\over 6})\\) = \\(cos{\\pi\\over 4}\\)<\/p>\n
\\(\\implies\\) \\(\\theta – {\\pi\\over 6}\\) = \\(2n\\pi \\pm {\\pi\\over 4}\\), n \\(\\in\\) Z.<\/p>\n
\\(\\implies\\) \\(\\theta\\) = \\(2n\\pi \\pm {\\pi\\over 4} + {\\pi\\over 6}\\), n \\(\\in\\) Z.<\/p>\n
\\(\\implies\\) \\(\\theta\\) = \\(2n\\pi + {\\pi\\over 4} + {\\pi\\over 6}\\) or, \\(\\theta\\) = \\(2n\\pi – {\\pi\\over 4} + {\\pi\\over 6}\\)<\/p>\n
\\(\\implies\\) \\(\\theta\\) = \\(2n\\pi + {5\\pi\\over 12}\\) or \\(2n\\pi – {\\pi\\over 12}\\)<\/p>\n
Hence, \\(\\theta\\) = \\(2n\\pi + {5\\pi\\over 12}\\) or, \\(\\theta\\) = \\(2n\\pi – {5\\pi\\over 12}\\), where n \\(\\in\\) Z.<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : We have, \\(\\sqrt{3} cos \\theta\\) + \\(sin \\theta\\) = \\(\\sqrt{2}\\) ………….(i) This is of the form \\(a cos\\theta\\) + \\(b sin \\theta\\) = c, where a = \\(\\sqrt{3}\\), b = 1 and c = \\(\\sqrt{2}\\). Let a = \\(r cos\\alpha\\) and b = \\(r sin\\alpha\\). Then, \\(\\sqrt{3}\\) = …<\/p>\n