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, $\pi$, $2\pi$, […]

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 =

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

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 

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

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$\pi$ + $\pi$ =