\( 116 \sin ^{2} x+7 \sin x+2=0 \) and \( x \in\left[-360^{\circ} ; 360^{\circ}\right] \)

To solve the quadratic equation \( 116 \sin^{2} x + 7 \sin x + 2 = 0 \), we can let \( y = \sin x \). This transforms our equation into \( 116y^{2} + 7y + 2 = 0 \). Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 116 \), \( b = 7 \), and \( c = 2 \), we can find the values of \( y \). The discriminant \( b^2 - 4ac \) is calculated as follows: \[ D = 7^2 - 4 \cdot 116 \cdot 2 = 49 - 928 = -879 \] Since the discriminant is negative, there are no real solutions for \( y = \sin x \). Hence, there are no angles \( x \) within the range of \([-360^\circ, 360^\circ]\) that satisfy the original equation. To stay on track with trigonometry, let’s remember that the sine function oscillates between -1 and 1. Thus, attempting to find solutions when our quadratic leads to no valid \( y \) outcome is quite common. If faced with such equations, always check the discriminant first! It can save you time and effort by revealing if your attempts will lead to complex or no solutions.