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General Solution of Trigonometric Equation

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Question

$Solve the following equations: (i) t a n θ + t a n 2 θ + t a n 3 θ = 0 (i i) t a n θ + t a n 2 θ = t a n 3 θ (i i i) t a n 3 θ + t a n θ = 2 t a n 2 θ$

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Solution

$(i) t a n θ + t a n 2 θ + t a n 3 θ = 0 t a n x + t a n 2 x + \frac{(t a n x + t a n 2 x)}{1 - t a n x . t a n 2 x} = 0 [t a n x + t a n 2 x] [\begin{matrix} 1 + \frac{1}{1 - t a n x . t a n 2 x} \end{matrix}] = 0 t a n x + t a n 2 x (2 - t a n x . t a n 2 x) = 0 t a n x = t a n (- 2 x) o r t a n x . t a n 2 x = 2 x = n π - 2 x O r t a n x . \frac{2 t a n x}{1 - t a n^{2} x} = 2 x = n π O r \frac{2 t a n x}{1 - t a n^{2} x} = 2 3 x = n π O r 2 t a n^{2} x = 2 - 2 t a n^{2} 3 x = n π O r 4 t a n^{2} x = 2 x = \frac{n π}{3} O r t a n^{2} x = 1 / 2 x = \frac{n π}{3} O r x = m π \pm t a n^{-} 1 (\frac{1}{\sqrt{2}}), n, m \in Z (i i) t a n θ + t a n 2 θ = t a n 3 θ t a n θ + t a n 2 θ = t a n (θ + 2 θ) t a n θ + t a n 2 θ - \frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = 0 [t a n θ + t a n 2 θ] [\begin{matrix} 1 - \frac{1}{1 - t a n θ t a n 2 θ} \end{matrix}] = 0 [t a n θ + t a n 2 θ] [\begin{matrix} \frac{1 - t a n θ t a n 2 θ - 1}{1 - t a n θ t a n 2 θ} \end{matrix}] = 0 [t a n θ + t a n 2 θ] [\begin{matrix} \frac{- t a n θ t a n 2 θ}{1 - t a n θ t a n 2 θ} \end{matrix}] = 0 t a n θ = 0 o r t a n 2 θ = 0 o r t a n θ + t a n 2 θ = 0 θ = n π o r \frac{n π}{2} o r t a n θ [\begin{matrix} \frac{1 - t a n^{2} θ + 2}{1 - t a n^{2} θ} \end{matrix}] = 0 θ = n π o r \frac{n π}{2} o r t a n θ = \pm \sqrt{3} θ = n π o r \frac{n π}{3} m, i, \in Z (i i i) t a n 3 θ + t a n θ = 2 t a n 2 θ We have, t a n 2 θ + t a n θ = 2 t a n 2 θ \Rightarrow t a n 3 θ - t a n 2 θ = t a n 2 θ - t a n θ \Rightarrow t a n 3 θ - t a n 2 θ = t a n 2 θ - t a n θ \Rightarrow 2 s i n^{2} θ s i n 2 θ = 0 \Rightarrow either s i n θ = 0 O r s i n 2 θ = 0 \Rightarrow θ = n π, n \in Z O r 2 θ = m π, m \in Z \Rightarrow θ = n π, n \in Z O r θ = m \frac{π}{2}, m \in Z$

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General Solution of Trigonometric Equation

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