Trigonometric Equations Questions

Solution : The general solution of \(sin \theta\) = 0 is given by \(\theta\) = \(n\pi\), n \(\in\) Z. Proof : We have, \(sin \theta\) = \(PM\over OP\) \(\therefore\)   \(sin \theta\) = 0 \(\implies\)  \(PM\over OP\) = 0 \(\implies\) PM = 0 \(\implies\)  OP coincides with OX or OX’ \(\implies\)  \(\theta\) = 0, \(\pi\), \(2\pi\), …

What is the General Solution of \(Sin \theta\) = 0 ? Read More »

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 = …

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

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 »

Solution : we have, cos3x + (sin2x – sin4x) = 0 = cos3x – 2sinx.cos3x = 0 \(\implies\)  (cos3x)(1 – 2sinx) = 0 \(\implies\)  cos3x = 0  or  sinx = \(1\over 2\) \(\implies\)  cos3x = 0 = cos\(\pi\over 2\)  or  sinx = \(1\over 2\) = sin\(\pi\over 6\) \(\implies\)  3x = 2n\(\pi\) \(\pm\) \(\pi\over 2\)  or  …

Solve : cos3x + sin2x – sin4x = 0 Read More »

Solution : we have, 6 – 10cosx = 3\(sin^2x\) \(\therefore\)  6 – 10cosx = 3 – 3\(cos^2x\) \(\implies\)  3\(cos^2x\) – 10cosx + 3 = 0 \(\implies\)  (3cosx-1)(cosx-3) = 0  \(\implies\)  cosx = \(1\over 3\) or cosx = 3 Since cosx = 3 is not possible as -1 \(\le\) cosx \(\le\) 1 \(\therefore\)  cosx = \(1\over …

Solve : 6 – 10cosx = 3\(sin^2x\) Read More »

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\(\pi\) + \(\pi\) = …

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