Trigonometric Equations Questions

Find the number of solutions of tanx + secx = 2cosx in [0, \(2\pi\)].

Solution : Here, tanx + secx = 2cosx       \(\implies\)     sinx + 1 = \(2cos^2x\) \(\implies\) \(2sin^2x\) + sinx – 1 = 0     \(\implies\)    sinx = \(1\over 2\), -1 But sinx = -1 \(\implies\) x = \(3\pi\over 2\) for which tanx + secx = 2cosx  is not defined. Thus, sinx = …

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If \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. then the general solution for \(\theta\) is

Solution : Since, \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. \(\implies\) \(cos^2\theta\) = \({1\over 6}sin\theta\).\(cos\theta\) \(\implies\) \(6cos^3\theta\) + \(cos^2\theta\) – 1 = 0 \(\therefore\)   (\(2cos\theta – 1\))(\(3cos^2\theta\) + \(2cos\theta\) + 1) = 0 \(cos\theta\) = \(1\over 2\)       (other values are imaginary) \(cos\theta\) = \(cos\pi\over 3\)   \(\theta\) = \(2n\pi \pm {\pi\over 3}\),  n …

If \({1\over 6}sin\theta\), \(cos\theta\) and \(tan\theta\) are in G.P. then the general solution for \(\theta\) is Read More »

Find general solution of (2sinx – cosx)(1 + cosx) = \(sin^2x\)

Solution : (2sinx – cosx)(1 + cosx) – (1 – \(cos^2x\)) = 0 \(\therefore\) (1 + cosx)(2sinx – cosx – 1 + cosx) = 0 \(\therefore\)  (1 + cosx)(2sinx – 1) = 0 \(\implies\) cosx = -1  or  sinx = \(1\over 2\) \(\implies\)  cosx = -1 = cos\(\pi\)  \(\implies\)  x = 2n\(\pi\) + \(\pi\) = …

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