Solve $\newline$ $9 \sin ^{2}\left(2 x-\frac{\pi}{4}\right)+\sin \left(4 x-\frac{\pi}{2}\right)=8$ $\newline$ in the interval $0<x<\frac{\pi}{2}$ $\newline$ Give your answers to one decimal place. $\newline$ [ $2$ mark

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Q. Solve

\newline

9 \sin ^{2}\left(2 x-\frac{\pi}{4}\right)+\sin \left(4 x-\frac{\pi}{2}\right)=8

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in the interval

0<x<\frac{\pi}{2}

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Give your answers to one decimal place.

\newline

[

2

mark

Rewrite Equation: First, let's rewrite the given equation in a more readable form: $\newline$ $9\sin^2(2x - \frac{\pi}{4}) + \sin(4x - \frac{\pi}{2}) = 8$ $\newline$ We need to solve this trigonometric equation in the interval $0 < x < \frac{\pi}{2}$ .
Use Trigonometric Identities: We can use the double-angle identity for sine, which states that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ . However, we need to be careful because the equation involves $\sin^2$ and sin of a double angle, which is not directly the double-angle identity.
Substitute Values: Let's first focus on the second term $\sin(4x - \frac{\pi}{2})$ . We can use the angle subtraction identity for sine, which states that $\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$ . In this case, $a = 4x$ and $b = \frac{\pi}{2}$ , so we have: $\newline$ $\sin(4x - \frac{\pi}{2}) = \sin(4x)\cos(\frac{\pi}{2}) - \cos(4x)\sin(\frac{\pi}{2})$ $\newline$ Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$ , this simplifies to: $\newline$ $\sin(4x - \frac{\pi}{2}) = -\cos(4x)$
Apply Double-Angle Identity: Now, we can substitute $-\cos(4x)$ back into the original equation: $9\sin^2(2x - \frac{\pi}{4}) - \cos(4x) = 8$
Simplify Equation: Next, we can use the double-angle identity for cosine, which states that $\cos(2\theta) = 1 - 2\sin^2(\theta)$ . We can apply this to $\cos(4x)$ by considering $4x$ as $2(2x)$ : $\cos(4x) = 1 - 2\sin^2(2x)$
Combine $\sin^2$ Terms: Substitute $\cos(4x)$ with $1 - 2\sin^2(2x)$ in the equation: $\newline$ $9\sin^2(2x - \frac{\pi}{4}) - (1 - 2\sin^2(2x)) = 8$
Isolate $\sin^2$ : Now, let's simplify the equation by distributing the negative sign and combining like terms: $\newline$ $9\sin^2(2x - \frac{\pi}{4}) - 1 + 2\sin^2(2x) = 8$ $\newline$ $9\sin^2(2x - \frac{\pi}{4}) + 2\sin^2(2x) = 9$
Solve for Sin: Combine the $\sin^2$ terms: $\newline$ $(9 + 2)\sin^2(2x - \frac{\pi}{4}) = 9$ $\newline$ $11\sin^2(2x - \frac{\pi}{4}) = 9$
Find Angle: Divide both sides by $11$ to isolate $\sin^2(2x - \frac{\pi}{4})$ : $\sin^2(2x - \frac{\pi}{4}) = \frac{9}{11}$
Calculate $x$ : Take the square root of both sides to solve for $\sin(2x - \pi/4)$ :
$\sin(2x - \pi/4) = \pm\sqrt{9/11}$
Since we are looking for solutions in the interval $0 < x < \pi/2$ , we only consider the positive square root because $\sin(2x - \pi/4)$ will be positive in this interval.
$\sin(2x - \pi/4) = \sqrt{9/11}$
Round Answer: Now, we need to find the angle whose sine is $\sqrt{\frac{9}{11}}$ . We can use a calculator to find this angle: $\newline$ $\sin(2x - \frac{\pi}{4}) = \sqrt{\frac{9}{11}} \approx 0.906$ $\newline$ Using the inverse sine function, we find: $\newline$ $2x - \frac{\pi}{4} \approx \arcsin(0.906)$
Round Answer: Now, we need to find the angle whose sine is $\sqrt{\frac{9}{11}}$ . We can use a calculator to find this angle: $\newline$ $\sin(2x - \frac{\pi}{4}) = \sqrt{\frac{9}{11}} \approx 0.906$ $\newline$ Using the inverse sine function, we find: $\newline$ $2x - \frac{\pi}{4} \approx \arcsin(0.906)$ Using a calculator to find the $\arcsin(0.906)$ , we get: $\newline$ $2x - \frac{\pi}{4} \approx 1.119$ (to three decimal places)
Round Answer: Now, we need to find the angle whose sine is $\sqrt{\frac{9}{11}}$ . We can use a calculator to find this angle: $\newline$ $\sin(2x - \frac{\pi}{4}) = \sqrt{\frac{9}{11}} \approx 0.906$ $\newline$ Using the inverse sine function, we find: $\newline$ $2x - \frac{\pi}{4} \approx \arcsin(0.906)$ Using a calculator to find the $\arcsin(0.906)$ , we get: $\newline$ $2x - \frac{\pi}{4} \approx 1.119$ (to three decimal places)Now, we solve for $x$ : $\newline$ $2x \approx 1.119 + \frac{\pi}{4}$ $\newline$ $2x \approx 1.119 + 0.785$ (since $\frac{\pi}{4} \approx 0.785$ ) $\newline$ $2x \approx 1.904$ $\newline$ $x \approx \frac{1.904}{2}$ $\newline$ $x \approx 0.952$
Round Answer: Now, we need to find the angle whose sine is $\sqrt{9/11}$ . We can use a calculator to find this angle: $\newline$ $\sin(2x - \pi/4) = \sqrt{9/11} \approx 0.906$ $\newline$ Using the inverse sine function, we find: $\newline$ $2x - \pi/4 \approx \arcsin(0.906)$ Using a calculator to find the $\arcsin(0.906)$ , we get: $\newline$ $2x - \pi/4 \approx 1.119$ (to three decimal places) Now, we solve for $x$ : $\newline$ $2x \approx 1.119 + \pi/4$ $\newline$ $2x \approx 1.119 + 0.785$ (since $\pi/4 \approx 0.785$ ) $\newline$ $2x \approx 1.904$ $\newline$ $\sin(2x - \pi/4) = \sqrt{9/11} \approx 0.906$ $0$ $\newline$ $\sin(2x - \pi/4) = \sqrt{9/11} \approx 0.906$ $1$ We round our answer to one decimal place as requested: $\newline$ $\sin(2x - \pi/4) = \sqrt{9/11} \approx 0.906$ $2$