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Arrange the steps in the correct order to solve this trigonometric equation:

[tex]\[
2 \sin^2 x - \sin x - 1 = 0 \text{ for } 0^{\circ} \leq x \leq 90^{\circ}
\][/tex]

1. Substitute [tex]\(\sin x = u\)[/tex]
2. [tex]\(2u^2 - u - 1 = 0\)[/tex]
3. Factor: [tex]\((2u + 1)(u - 1) = 0\)[/tex]
4. Solve for [tex]\(u\)[/tex]: [tex]\(u = -\frac{1}{2}\)[/tex] and [tex]\(u = 1\)[/tex]
5. Replace [tex]\(u\)[/tex] with [tex]\(\sin x\)[/tex]: [tex]\(\sin x = -\frac{1}{2}\)[/tex] and [tex]\(\sin x = 1\)[/tex]
6. Find [tex]\(x\)[/tex]: [tex]\(x = 90^{\circ}\)[/tex] (within the given range)



Answer :

GinnyAnswer
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.

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