Question

Let ( sin x=t )
( 32^{3} t^{2}-7 z+2=0 )
( 2 y quad 3 z^{2}-6 z-t+2=0 )
2) ( (3 z-1)(z-2)=0 )
( Rightarrow quad t=frac{1}{3}, 2 )
( therefore sin x=frac{1}{3} ) a ( sin x=2 x_{text {becaun }} ) sinn 31
( sin n=frac{1}{3} )
( therefore x=n pi+(-1)^{n} sin ^{-1} frac{1}{3} quad(0,2 pi) quad(2 pi, 4 pi) mid 4 pi 9 )
( therefore ) no. of ( operatorname{seln} i n[0,5 pi]=2+2+2 )
( =6 )
give rating

# 7. The The number of values of x in the interval [0, 51] satisfying the equation 3 sin2 x - 7 sin x + 2 = 0 is (B) 5 (A) O

Solution