Question

find x where 0 ≤ x ≤ 2π. 2 sin²x + 2 = 5 sinx. (π/?), ( π/ ). enter the smaller answer first.

Explanation:

Step1: Rearrange the equation

Let \( y = \sin x \), then the equation \( 2\sin^{2}x + 2 = 5\sin x \) becomes \( 2y^{2}-5y + 2 = 0 \).

Step2: Solve the quadratic equation

For a quadratic equation \( ay^{2}+by + c = 0 \) (here \( a = 2 \), \( b=- 5 \), \( c = 2 \)), the quadratic formula is \( y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-5)^{2}-4\times2\times2 = 25 - 16=9 \).
Then \( y=\frac{5\pm\sqrt{9}}{4}=\frac{5\pm3}{4} \).
We get two solutions:

• When we take the plus sign: \( y=\frac{5 + 3}{4}=\frac{8}{4}=2 \). But \( - 1\leqslant\sin x\leqslant1 \), so \( y = 2 \) is not valid.
• When we take the minus sign: \( y=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2} \).

Step3: Find x from \( \sin x=\frac{1}{2} \)

We know that \( \sin x=\frac{1}{2} \) and \( 0\leqslant x\leqslant2\pi \).
The solutions for \( x \) are \( x=\frac{\pi}{6} \) (in the first quadrant) and \( x=\frac{5\pi}{6} \) (in the second quadrant).

Answer:

The smaller answer is \( \frac{\pi}{6} \) (so the first box is 6) and the other is \( \frac{5\pi}{6} \) (the second box numerator is 5, denominator is 6). So filling the boxes: first box: 6; second box numerator: 5, denominator: 6. The solutions for \( x \) are \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \).