Let's arrange the steps in the correct order to solve the trigonometric equation [tex]\(2 \sin^2 x - \sin x - 1 = 0\)[/tex] for [tex]\(0^\circ \leq x \leq 90^\circ\)[/tex].
1. Use a substitution, let [tex]\( u = \sin x \)[/tex].
2. Replace [tex]\( \sin x \)[/tex] with [tex]\( u \)[/tex] in the equation: [tex]\( 2 u^2 - u - 1 = 0 \)[/tex].
3. Factor the quadratic equation: [tex]\((2 u + 1)(u - 1) = 0\)[/tex].
4. Solve for [tex]\( u \)[/tex]: [tex]\( u = 1 \)[/tex] and [tex]\( u = -\frac{1}{2} \)[/tex].
5. Replace [tex]\( u \)[/tex] with [tex]\( \sin x \)[/tex]: [tex]\(\sin x = 1 \)[/tex] or [tex]\(\sin x = -\frac{1}{2} \)[/tex].
6. Solve for [tex]\( x \)[/tex] in the given range [tex]\( 0^\circ \leq x \leq 90^\circ \)[/tex]: [tex]\( x = 90^\circ \)[/tex] (since [tex]\(\sin x = -\frac{1}{2}\)[/tex] is not possible within this range).
So, the correct arrangement of steps is:
1. [tex]\( 2 \sin^2 x - \sin x - 1 = 0 \text{ for } 0^\circ \leq x \leq 90^\circ \)[/tex]
2. Use a substitution, let [tex]\( u = \sin x \)[/tex].
3. Replace [tex]\( \sin x \)[/tex] with [tex]\( u \)[/tex] in the equation: [tex]\( 2 u^2 - u - 1 = 0 \)[/tex].
4. Factor the quadratic equation: [tex]\((2 u + 1)(u - 1) = 0\)[/tex].
5. Solve for [tex]\( u \)[/tex]: [tex]\( u = 1 \)[/tex] and [tex]\( u = -\frac{1}{2} \)[/tex].
6. Replace [tex]\( u \)[/tex] with [tex]\( \sin x \)[/tex]: [tex]\(\sin x = 1 \)[/tex] or [tex]\(\sin x = -\frac{1}{2} \)[/tex].
7. Solve for [tex]\( x \)[/tex] in the given range [tex]\( 0^\circ \leq x \leq 90^\circ \)[/tex]: [tex]\( x = 90^\circ \)[/tex].
