Here is the correct order to solve the trigonometric equation for [tex]\( 2 \sin^2 x - \sin x - 1 = 0 \)[/tex] within the interval [tex]\(0^\circ \leq x \leq 90^\circ\)[/tex]:
1. [tex]\(2 \sin^2 x-\sin x-1=0\)[/tex] 2. substituting [tex]\(\sin x=u\)[/tex] 3. [tex]\(2 u^2-u-1=0\)[/tex] 4. [tex]\( (2u+1)(u-1)=0 \)[/tex] 5. [tex]\( u=1 \)[/tex] and [tex]\( u=-\frac{1}{2} \)[/tex] 6. replacing [tex]\(u\)[/tex] with [tex]\(\sin x\)[/tex] : [tex]\(\sin x=1\)[/tex] or [tex]\(\sin x=-\frac{1}{2}\)[/tex] 7. [tex]\(x=90^{\circ} \text { for } 0^{\circ} \leq x \leq 90^{\circ} \)[/tex]
Here is the step-by-step solution in clear and logical order.
