हल कीजिए -
(i) ` sin 3 theta = sin 2 theta`
(ii) ` sin m theta + sin n theta = 0`
(iii) ` tan 3 theta tan theta = 1`

Explanation

दिया है - `sin 3 theta = sin 2 theta`
`= sin [n pi + (-1)^(n) * 2 theta]`
`rArr " " 3 theta = n pi + (-1)^(n) * 2 theta `
यदि n सम है अर्थात n = 2m तब
`3 theta = 2 m pi + 2 theta`
`rArr theta = 2 m pi`
और यदि n विषम संख्या है अर्थात `n = 2m +1`, तब
`rArr 3 theta = (2m +1) pi - 2 theta`
`rArr 5 theta = (2m +1) pi`
`rArr theta = (2m +1) * pi/5`
(ii) दिया है - `sin m theta + sin n theta = 0`
`2 sin [((m+n)theta)/2] cos [((m-n)theta)/2]= 0`
यदि `sin((m+n)/2) theta = 0 rArr ((m+n)/2) theta = k pi, " " k in Z`
`rArr theta = (2kpi)/(m+n), " " k in Z`
यदि `cos((m-n)/2) theta = 0 `तब `((m-n)/2) theta = (2k+1) pi/2, " " k in Z`
`rArr theta = ((2k+1)pi)/(m-n) , " " k in Z`
(iii) दिया है - `tan 3 theta tan theta =1`
`rArr tan 3 theta = 1/(tan theta) = cot theta = tan (pi/2 - theta)`
`rArr 3 theta = n pi + (pi/2 - theta) `
`rArr 4 theta = (2 pi +1) pi/2`
` rArr theta = (2n +1) pi/8`