Question

$\text{Solve the following equations :}\left(i\right)cot\theta +tan\theta =2\left(ii\right)2si{n}^2\theta =3cos\theta ,0\le \theta \le 2\pi \left(iii\right)sec\theta cos5\theta +1=0,0<\theta <\frac{\pi }{2}\left(iv\right)5co{s}^2\theta +7si{n}^2\theta -6=0\left(v\right)sinx-3sin2x+sin3x=cosx-3cos2x+cos3x$

Answer 1

$\left(i\right)cot\theta +tan\theta =2\frac{cos\theta }{sin\theta }+\frac{sin\theta }{cos\theta }=2\frac{co{s}^2\theta +si{n}^2\theta }{sin\theta cos\theta }=2si{n}^2\theta +co{s}^2\theta =2sin\theta cos\theta 1=2sin\theta cos\theta [\because si{n}^2\theta +co{s}^2\theta =1]2sin\theta cos\theta =1sin2\theta =1sin2\theta =sin\frac{\pi }{2}sin2\theta =sin\frac{(2n+1)}{2}\pi 2\theta =\frac{(2n-1)}{2}\pi x=\frac{(2n+1)}{4}\pi ,n\in Z\left(ii\right)2si{n}^2\theta =3cos\theta ,0\le \theta \le 2\pi 2-2co{s}^2\theta =3cos\theta 2co{s}^2\theta +3cos\theta -2=02co{s}^2\theta +4cos\theta -cos\theta -2=0(cos\theta +2)(2cos\theta -1)=0cos\theta =-2orcos\theta =0.5cos\theta =-2,neverpossiblecos\theta =0.5=60,300\left(iii\right)sec\theta cos5\theta +1=0,0<\theta <\frac{\pi }{2}sec\theta cos5\theta +1=0\frac{cos5x+cosx}{cosx}=0\Rightarrow cosx\ne 02cos3xcos2x=0cos3x=0orcos2x=03x=\frac{\pi }{2}or2x=\frac{\pi }{2}x=\frac{\pi }{4},\frac{\pi }{6}\left(iv\right)5co{s}^2\theta +7si{n}^2\theta -6=05(1-si{n}^2\theta )+7si{n}^2\theta -6=0[Ascos2\theta =1-si{n}^2\theta ]5-si{n}^2\theta +7si{n}^2\theta -6=05+2si{n}^2\theta -6=02si{n}^2\theta -1=02si{n}^2\theta =1si{n}^2\theta =\frac{1}{2}sin\theta =\pm \frac{1}{\sqrt{2}}\theta =n\pi \pm \frac{\pi }{4}\left(v\right)sinx-3sin2x+sin3x=cosx-3cos2x+cos3x(sinx+sin3x)-3sin2x-(cosx+cos3x)+3cos2x=02sin2xcosx-3sin2x-2cos2xcosx+2cos2x=0sin2x(2cosx-3)-cos2x(2cosx-3)=0(2cosx-3)(sin2x-cos2x)=0cosx=\frac{3}{2}orsin2x-cos2x=0butcosx\in [-1,1]\Rightarrow cosx\ne \frac{3}{2}sin2x=cos2xtan2x=12x=n\pi +\frac{\pi }{4}x=\frac{n\pi }{2}+\frac{\pi }{8}$