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The equations kcosx - 3sinx=k+1 is solvable only if k belongs to the interval?
k(cos x) - 3(sin x) = k + 1k(cos x) - (k + 1) = 3(sin x)[k(cos x) - (k + 1)]^2 = [3(sin x)]^2k^2cos^2x - 2k(k + 1)(cos x) + (k + 1)^2 = 9sin^2xk^2cos^2x - (2k^2 + 2k)(cos x) + (k + 1)^2 = 9(1 - cos^2x)(k^2 + 9)cos^2x - (2k^2 + 2k)(cos x) + (k^2 + 2k - 8) = 0Here we have a quadratic equation in terms of (cos x).This equation is only solvable when the determinant b² - 4ac greater than or equal to 0, otherwise the solution will be complex.b² - 4ac greater than or equal to 0[-(2k^2 + 2k)]^2 - 4(k^2 + 9)(k^2 + 2k - 8) greater than or equal to 0(4k^4 + 8k^3 + 4k^2) - (4k^4 + 8k^3 + 4k^2 + 72k - 288) greater than or equal to 0-72k + 288 greater than or equal to 072k less than or equal to 288k less than or equal to 4i.e. -infinity<k<4 = k belongs to (-infinity, 4]
k(cos x) - 3(sin x) = k + 1k(cos x) - (k + 1) = 3(sin x)[k(cos x) - (k + 1)]^2 = [3(sin x)]^2k^2cos^2x - 2k(k + 1)(cos x) + (k + 1)^2 = 9sin^2xk^2cos^2x - (2k^2 + 2k)(cos x) + (k + 1)^2 = 9(1 - cos^2x)(k^2 + 9)cos^2x - (2k^2 + 2k)(cos x) + (k^2 + 2k - 8) = 0Here we have a quadratic equation in terms of (cos x).This equation is only solvable when the determinant b² - 4ac greater than or equal to 0, otherwise the solution will be complex.b² - 4ac greater than or equal to 0[-(2k^2 + 2k)]^2 - 4(k^2 + 9)(k^2 + 2k - 8) greater than or equal to 0(4k^4 + 8k^3 + 4k^2) - (4k^4 + 8k^3 + 4k^2 + 72k - 288) greater than or equal to 0-72k + 288 greater than or equal to 072k less than or equal to 288k less than or equal to 4i.e. -infinity<k<4
= k belongs to (-infinity, 4]
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