Question

The equation $4{\sin }^{2}x+4\sin x+a^2-3=0$ passesses a solution if ${}^{\prime }{a}^{\prime }$ belongs to the interval

• A. $(-1,3)$
• B. $(-3,1)$
• C. $(-2,2)$
• D. $R-(-2,2)$

Answer 1

The correct option is D $R-(-2,2)$

Given $4{\sin }^{2}x+4\sin x+a^2-3=0$

$(2\sin x{)}^2+2(2\sin x\left)\right(1)+1^2-1^2+a^2-3=0$

$(2\sin x+1{)}^2+a^2-4=0$

$⟹(2\sin x+1{)}^2=(4-a^2)$

As $(2\sin x+1{)}^2\ge 0$

$⟹4-a^2\ge 0⟹a\in [-2,2]⟹a\in R-(-2,2)$