`sin^(2) x+1/4 sin^(2) 3x= sin x sin^(2) 3x` or `sin^(2) x-sin x sin^(2) 3x+1/4 sin^(2) 3x=0` or `(sin x-1/2 sin^(2) 3x)^(2) +1/4 sin^(2) 3x (1- sin^(2) 3x)=0` or `(sin x-1/2 sin^(2) 3x)^(2) +1/4 sin^(2) 3x cos^(2) 3x=0` or `(sin x 1/2 sin^(2) 3x)^(2) +1/16 sin^(2) 6x =0` or `sin x-1/2 sin^(2) 3x=0` and `sin 6x=0` or `2 sin x=sin^(2) 3x and sin 6x =0` From `sin 6x=0, x=kpi//6, k in Z` From here, we choose those values which satisfy the equation `2 sin x=sin^(2) 3x`. Now, `sin^(2) 3 ((k pi)/6)=sin^(2) (kpi)/2={("1, if k is odd"),("0, if k is even"):}}`
Topper's Solved these Questions
TRIGONOMETRIC EQUATIONS
CENGAGE|Exercise Exercise 4.1|12 Videos
TRIGONOMETRIC EQUATIONS
CENGAGE|Exercise Exercise 4.2|6 Videos
TRIGNOMETRIC RATIOS IDENTITIES AND TRIGNOMETRIC EQUATIONS
CENGAGE|Exercise Question Bank|34 Videos
TRIGONOMETRIC FUNCTIONS
CENGAGE|Exercise SINGLE CORRECT ANSWER TYPE|38 Videos