Answer :
Sure, let's break down the solution for each part of the problem step-by-step.
### Part (a)
We need to show that the equation:
[tex]\[ 5 \sin x - \cos^2 x + 2 \sin^2 x = 1 \][/tex]
can be written in the form:
[tex]\[ 3 \sin^2 x + 5 \sin x - 2 = 0 \][/tex]
First, recall the Pythagorean identity:
[tex]\[ \cos^2 x = 1 - \sin^2 x \][/tex]
Substitute [tex]\(\cos^2 x\)[/tex] in the given equation:
[tex]\[ 5 \sin x - (1 - \sin^2 x) + 2 \sin^2 x = 1 \][/tex]
Now, distribute and combine like terms:
[tex]\[ 5 \sin x - 1 + \sin^2 x + 2 \sin^2 x = 1 \][/tex]
[tex]\[ 5 \sin x - 1 + 3 \sin^2 x = 1 \][/tex]
Finally, move -1 to the right side of the equation:
[tex]\[ 3 \sin^2 x + 5 \sin x - 2 = 0 \][/tex]
Thus, we have shown that the equation can be written in the required form.
### Part (b)
To solve the equation for [tex]\(\theta\)[/tex] in the interval [tex]\(-180^\circ \leqslant \theta < 180^\circ\)[/tex], we start with the equation:
[tex]\[ 5 \sin (2\theta) - \cos^2 (2\theta) + 2 \sin^2 (2\theta) = 1 \][/tex]
Using the identity [tex]\(\cos^2 (2\theta) = 1 - \sin^2 (2\theta)\)[/tex], we can substitute [tex]\(\cos^2 (2\theta)\)[/tex]:
[tex]\[ 5 \sin (2\theta) - (1 - \sin^2 (2\theta)) + 2 \sin^2 (2\theta) = 1 \][/tex]
Now simplify and combine like terms:
[tex]\[ 5 \sin (2\theta) - 1 + \sin^2 (2\theta) + 2 \sin^2 (2\theta) = 1 \][/tex]
[tex]\[ 5 \sin (2\theta) - 1 + 3 \sin^2 (2\theta) = 1 \][/tex]
Move constants to the right side:
[tex]\[ 3 \sin^2 (2\theta) + 5 \sin (2\theta) - 2 = 0 \][/tex]
Let [tex]\( y = \sin (2\theta) \)[/tex]. Now we solve the quadratic equation:
[tex]\[ 3y^2 + 5y - 2 = 0 \][/tex]
Solve for [tex]\( y \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{25 + 24}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{49}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm 7}{6} \][/tex]
So the solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
[tex]\[ y = \frac{-5 - 7}{6} = \frac{-12}{6} = -2 \][/tex]
Since [tex]\( y = \sin(2\theta) \)[/tex], we have [tex]\( \sin(2\theta) = \frac{1}{3} \)[/tex] and [tex]\( \sin(2\theta) = -2 \)[/tex]. The second case [tex]\( \sin(2\theta) = -2 \)[/tex] is not possible because the sine function ranges from -1 to 1. Therefore, we discard [tex]\( \sin(2\theta) = -2 \)[/tex].
Thus, we are left with:
[tex]\[ \sin(2\theta) = \frac{1}{3} \][/tex]
Now solve for [tex]\( 2\theta \)[/tex]:
[tex]\[ 2\theta = \arcsin\left(\frac{1}{3}\right) \][/tex]
We need solutions for [tex]\( 2\theta \)[/tex] in the range [tex]\(-360^\circ \leqslant 2\theta < 360^\circ\)[/tex]. The principal value is:
[tex]\[ 2\theta = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ \][/tex]
Since sine is also positive in the second quadrant:
[tex]\[ 2\theta = 180^\circ - 19.47^\circ = 160.53^\circ \][/tex]
So the solutions for [tex]\( 2\theta \)[/tex] in the given range are:
[tex]\[ 2\theta = 19.47^\circ, 160.53^\circ, 19.47^\circ - 360^\circ = -340.53^\circ, 160.53^\circ - 360^\circ = -199.47^\circ \][/tex]
Now divide by 2 to find [tex]\( \theta \)[/tex]:
[tex]\[ \theta = \frac{19.47^\circ}{2} \approx 9.74^\circ \][/tex]
[tex]\[ \theta = \frac{160.53^\circ}{2} \approx 80.27^\circ \][/tex]
[tex]\[ \theta = \frac{-340.53^\circ}{2} \approx -170.27^\circ \][/tex]
[tex]\[ \theta = \frac{-199.47^\circ}{2} \approx -99.74^\circ \][/tex]
Therefore, the solutions for [tex]\(\theta\)[/tex] in the interval [tex]\(-180^\circ \leqslant \theta < 180^\circ\)[/tex] are:
[tex]\[ \theta \approx 9.74^\circ, 80.27^\circ, -170.27^\circ, -99.74^\circ \][/tex]
These are the angles to two decimal places.
### Part (a)
We need to show that the equation:
[tex]\[ 5 \sin x - \cos^2 x + 2 \sin^2 x = 1 \][/tex]
can be written in the form:
[tex]\[ 3 \sin^2 x + 5 \sin x - 2 = 0 \][/tex]
First, recall the Pythagorean identity:
[tex]\[ \cos^2 x = 1 - \sin^2 x \][/tex]
Substitute [tex]\(\cos^2 x\)[/tex] in the given equation:
[tex]\[ 5 \sin x - (1 - \sin^2 x) + 2 \sin^2 x = 1 \][/tex]
Now, distribute and combine like terms:
[tex]\[ 5 \sin x - 1 + \sin^2 x + 2 \sin^2 x = 1 \][/tex]
[tex]\[ 5 \sin x - 1 + 3 \sin^2 x = 1 \][/tex]
Finally, move -1 to the right side of the equation:
[tex]\[ 3 \sin^2 x + 5 \sin x - 2 = 0 \][/tex]
Thus, we have shown that the equation can be written in the required form.
### Part (b)
To solve the equation for [tex]\(\theta\)[/tex] in the interval [tex]\(-180^\circ \leqslant \theta < 180^\circ\)[/tex], we start with the equation:
[tex]\[ 5 \sin (2\theta) - \cos^2 (2\theta) + 2 \sin^2 (2\theta) = 1 \][/tex]
Using the identity [tex]\(\cos^2 (2\theta) = 1 - \sin^2 (2\theta)\)[/tex], we can substitute [tex]\(\cos^2 (2\theta)\)[/tex]:
[tex]\[ 5 \sin (2\theta) - (1 - \sin^2 (2\theta)) + 2 \sin^2 (2\theta) = 1 \][/tex]
Now simplify and combine like terms:
[tex]\[ 5 \sin (2\theta) - 1 + \sin^2 (2\theta) + 2 \sin^2 (2\theta) = 1 \][/tex]
[tex]\[ 5 \sin (2\theta) - 1 + 3 \sin^2 (2\theta) = 1 \][/tex]
Move constants to the right side:
[tex]\[ 3 \sin^2 (2\theta) + 5 \sin (2\theta) - 2 = 0 \][/tex]
Let [tex]\( y = \sin (2\theta) \)[/tex]. Now we solve the quadratic equation:
[tex]\[ 3y^2 + 5y - 2 = 0 \][/tex]
Solve for [tex]\( y \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{25 + 24}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{49}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm 7}{6} \][/tex]
So the solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
[tex]\[ y = \frac{-5 - 7}{6} = \frac{-12}{6} = -2 \][/tex]
Since [tex]\( y = \sin(2\theta) \)[/tex], we have [tex]\( \sin(2\theta) = \frac{1}{3} \)[/tex] and [tex]\( \sin(2\theta) = -2 \)[/tex]. The second case [tex]\( \sin(2\theta) = -2 \)[/tex] is not possible because the sine function ranges from -1 to 1. Therefore, we discard [tex]\( \sin(2\theta) = -2 \)[/tex].
Thus, we are left with:
[tex]\[ \sin(2\theta) = \frac{1}{3} \][/tex]
Now solve for [tex]\( 2\theta \)[/tex]:
[tex]\[ 2\theta = \arcsin\left(\frac{1}{3}\right) \][/tex]
We need solutions for [tex]\( 2\theta \)[/tex] in the range [tex]\(-360^\circ \leqslant 2\theta < 360^\circ\)[/tex]. The principal value is:
[tex]\[ 2\theta = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ \][/tex]
Since sine is also positive in the second quadrant:
[tex]\[ 2\theta = 180^\circ - 19.47^\circ = 160.53^\circ \][/tex]
So the solutions for [tex]\( 2\theta \)[/tex] in the given range are:
[tex]\[ 2\theta = 19.47^\circ, 160.53^\circ, 19.47^\circ - 360^\circ = -340.53^\circ, 160.53^\circ - 360^\circ = -199.47^\circ \][/tex]
Now divide by 2 to find [tex]\( \theta \)[/tex]:
[tex]\[ \theta = \frac{19.47^\circ}{2} \approx 9.74^\circ \][/tex]
[tex]\[ \theta = \frac{160.53^\circ}{2} \approx 80.27^\circ \][/tex]
[tex]\[ \theta = \frac{-340.53^\circ}{2} \approx -170.27^\circ \][/tex]
[tex]\[ \theta = \frac{-199.47^\circ}{2} \approx -99.74^\circ \][/tex]
Therefore, the solutions for [tex]\(\theta\)[/tex] in the interval [tex]\(-180^\circ \leqslant \theta < 180^\circ\)[/tex] are:
[tex]\[ \theta \approx 9.74^\circ, 80.27^\circ, -170.27^\circ, -99.74^\circ \][/tex]
These are the angles to two decimal places.
