हल कीजिए - ` sin 3 alpha = 4 sin alpha ⋅ sin ( θ + α ) ⋅ sin ( θ − α ) `

Explanation

` sin 3 alpha = 4 sin alpha* sin ( theta +alpha) * sin (theta-alpha)`
` rArr 3 sin alpha - 4 sin^(3) alpha - 4 sin^(3) alpha = 4 sin alpha* (sin^(2) theta- sin^(2)alpha)`
` rArr 3 - 4 sin^(2) alpha = 4 sin^(2) theta - 4 sin^(2) alpha`
` rArr 3 - 4 sin^(2) alpha = 4 sin^(2) theta - 4 sin^(2) alpha `
` rArr 4 sin^(2) theta = 3 `
` rArr sin^(2) theta = 3/4 = ((sqrt(3))/2)^(2)= sin^(2), pi/3 `
` rArr theta = n pi pm pi/3`