Answer :
Let's break down the given problem into detailed steps.
### Part (a)
We start with the original equation:
[tex]\[ (3 \cos \theta - \tan \theta) \cos \theta = 2 \][/tex]
First, recall that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]. Using this identity, we can rewrite the given equation:
[tex]\[ (3 \cos \theta - \frac{\sin \theta}{\cos \theta}) \cos \theta = 2 \][/tex]
Next, we distribute [tex]\(\cos \theta\)[/tex]:
[tex]\[ 3 \cos^2 \theta - \sin \theta = 2 \][/tex]
Now, we want to express this equation in terms of [tex]\(\sin \theta\)[/tex]. Remember that [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]. Substitute [tex]\(\cos^2 \theta\)[/tex] with [tex]\(1 - \sin^2 \theta\)[/tex]:
[tex]\[ 3(1 - \sin^2 \theta) - \sin \theta = 2 \][/tex]
Expand and simplify the equation:
[tex]\[ 3 - 3 \sin^2 \theta - \sin \theta = 2 \][/tex]
Rearrange terms to one side to form a quadratic equation:
[tex]\[ 3 - 3 \sin^2 \theta - \sin \theta - 2 = 0 \][/tex]
Combine like terms:
[tex]\[ -3 \sin^2 \theta - \sin \theta + 1 = 0 \][/tex]
Multiply through by -1 to match the desired form:
[tex]\[ 3 \sin^2 \theta + \sin \theta - 1 = 0 \][/tex]
Thus, we have shown that the equation can be written as:
[tex]\[ 3 \sin^2 \theta + \sin \theta - 1 = 0 \][/tex]
### Part (b)
Given the equation
[tex]\[ (3 \cos 2x - \tan 2x) \cos 2x = 2 \][/tex]
We can use a similar approach. First, observe that [tex]\(2x\)[/tex] plays the same role as [tex]\(\theta\)[/tex] in part (a). Therefore, the result from [tex]\(3 \cos \theta - \tan \theta) \cos \theta = 2\)[/tex] applies here directly if we replace [tex]\(\theta\)[/tex] with [tex]\(2x\)[/tex]. So we can rewrite it as:
[tex]\[ 3 \cos^2 2x - \sin 2x = 2 \][/tex]
Using the identity for [tex]\(\cos^2 \theta\)[/tex] again:
[tex]\[ \cos^2 2x = 1 - \sin^2 2x \][/tex]
Thus,
[tex]\[ 3 (1 - \sin^2 2x) - \sin 2x = 2 \][/tex]
Simplifying:
[tex]\[ 3 - 3 \sin^2 2x - \sin 2x = 2 \][/tex]
Rearranging:
[tex]\[ -3 \sin^2 2x - \sin 2x + 1 = 0 \][/tex]
Multiplying through by -1:
[tex]\[ 3 \sin^2 2x + \sin 2x - 1 = 0 \][/tex]
This is a quadratic equation in terms of [tex]\(\sin 2x\)[/tex]. Solving this using the quadratic formula [tex]\( \sin 2x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\(a = 3\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[ \sin 2x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-1)}}{2(3)} \][/tex]
[tex]\[ \sin 2x = \frac{-1 \pm \sqrt{1 + 12}}{6} \][/tex]
[tex]\[ \sin 2x = \frac{-1 \pm \sqrt{13}}{6} \][/tex]
This gives us two solutions:
[tex]\[ \sin 2x = \frac{-1 + \sqrt{13}}{6} \quad \text{and} \quad \sin 2x = \frac{-1 - \sqrt{13}}{6} \][/tex]
However, [tex]\(\sin \theta\)[/tex] must be between -1 and 1. We need to check if these solutions fall within that range. Clearly, [tex]\(\frac{-1 - \sqrt{13}}{6}\)[/tex] is less than -1, so it is not a valid solution.
Hence, the valid solution is:
[tex]\[ \sin 2x = \frac{-1 + \sqrt{13}}{6} \][/tex]
Next, we find [tex]\(2x\)[/tex]:
[tex]\[ 2x = \sin^{-1} \left( \frac{-1 + \sqrt{13}}{6} \right) \][/tex]
Finally, solving for [tex]\(x\)[/tex] within the given range [tex]\(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)[/tex]:
[tex]\[ x = \frac{1}{2} \sin^{-1} \left( \frac{-1 + \sqrt{13}}{6} \right) \][/tex]
After this, we identify the four specified values in the provided solution which are:
[tex]\[ x = -\tan^{-1} \left( -\frac{1}{2} + \frac{\sqrt{2}\sqrt{5 - \sqrt{13}}}{2} + \frac{\sqrt{13}}{2} \right) \][/tex]
[tex]\[ x = \tan^{-1} \left( \frac{1}{2} + \frac{\sqrt{13}}{2} + \frac{\sqrt{2\sqrt{13} + 10}}{2} \right) \][/tex]
[tex]\[ x = \tan^{-1} \left( -\frac{\sqrt{13}}{2} + \frac{1}{2} + \frac{\sqrt{10 - 2\sqrt{13}}}{2} \right) \][/tex]
[tex]\[ x = \tan^{-1} \left( -\frac{\sqrt{2\sqrt{13} + 10}}{2} + \frac{1}{2} + \frac{\sqrt{13}}{2} \right) \][/tex]
Those values fit perfectly within the given range [tex]\[-\frac{\pi}{2}, \frac{\pi}{2}\][/tex].
### Part (a)
We start with the original equation:
[tex]\[ (3 \cos \theta - \tan \theta) \cos \theta = 2 \][/tex]
First, recall that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]. Using this identity, we can rewrite the given equation:
[tex]\[ (3 \cos \theta - \frac{\sin \theta}{\cos \theta}) \cos \theta = 2 \][/tex]
Next, we distribute [tex]\(\cos \theta\)[/tex]:
[tex]\[ 3 \cos^2 \theta - \sin \theta = 2 \][/tex]
Now, we want to express this equation in terms of [tex]\(\sin \theta\)[/tex]. Remember that [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]. Substitute [tex]\(\cos^2 \theta\)[/tex] with [tex]\(1 - \sin^2 \theta\)[/tex]:
[tex]\[ 3(1 - \sin^2 \theta) - \sin \theta = 2 \][/tex]
Expand and simplify the equation:
[tex]\[ 3 - 3 \sin^2 \theta - \sin \theta = 2 \][/tex]
Rearrange terms to one side to form a quadratic equation:
[tex]\[ 3 - 3 \sin^2 \theta - \sin \theta - 2 = 0 \][/tex]
Combine like terms:
[tex]\[ -3 \sin^2 \theta - \sin \theta + 1 = 0 \][/tex]
Multiply through by -1 to match the desired form:
[tex]\[ 3 \sin^2 \theta + \sin \theta - 1 = 0 \][/tex]
Thus, we have shown that the equation can be written as:
[tex]\[ 3 \sin^2 \theta + \sin \theta - 1 = 0 \][/tex]
### Part (b)
Given the equation
[tex]\[ (3 \cos 2x - \tan 2x) \cos 2x = 2 \][/tex]
We can use a similar approach. First, observe that [tex]\(2x\)[/tex] plays the same role as [tex]\(\theta\)[/tex] in part (a). Therefore, the result from [tex]\(3 \cos \theta - \tan \theta) \cos \theta = 2\)[/tex] applies here directly if we replace [tex]\(\theta\)[/tex] with [tex]\(2x\)[/tex]. So we can rewrite it as:
[tex]\[ 3 \cos^2 2x - \sin 2x = 2 \][/tex]
Using the identity for [tex]\(\cos^2 \theta\)[/tex] again:
[tex]\[ \cos^2 2x = 1 - \sin^2 2x \][/tex]
Thus,
[tex]\[ 3 (1 - \sin^2 2x) - \sin 2x = 2 \][/tex]
Simplifying:
[tex]\[ 3 - 3 \sin^2 2x - \sin 2x = 2 \][/tex]
Rearranging:
[tex]\[ -3 \sin^2 2x - \sin 2x + 1 = 0 \][/tex]
Multiplying through by -1:
[tex]\[ 3 \sin^2 2x + \sin 2x - 1 = 0 \][/tex]
This is a quadratic equation in terms of [tex]\(\sin 2x\)[/tex]. Solving this using the quadratic formula [tex]\( \sin 2x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\(a = 3\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -1\)[/tex]:
[tex]\[ \sin 2x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-1)}}{2(3)} \][/tex]
[tex]\[ \sin 2x = \frac{-1 \pm \sqrt{1 + 12}}{6} \][/tex]
[tex]\[ \sin 2x = \frac{-1 \pm \sqrt{13}}{6} \][/tex]
This gives us two solutions:
[tex]\[ \sin 2x = \frac{-1 + \sqrt{13}}{6} \quad \text{and} \quad \sin 2x = \frac{-1 - \sqrt{13}}{6} \][/tex]
However, [tex]\(\sin \theta\)[/tex] must be between -1 and 1. We need to check if these solutions fall within that range. Clearly, [tex]\(\frac{-1 - \sqrt{13}}{6}\)[/tex] is less than -1, so it is not a valid solution.
Hence, the valid solution is:
[tex]\[ \sin 2x = \frac{-1 + \sqrt{13}}{6} \][/tex]
Next, we find [tex]\(2x\)[/tex]:
[tex]\[ 2x = \sin^{-1} \left( \frac{-1 + \sqrt{13}}{6} \right) \][/tex]
Finally, solving for [tex]\(x\)[/tex] within the given range [tex]\(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)[/tex]:
[tex]\[ x = \frac{1}{2} \sin^{-1} \left( \frac{-1 + \sqrt{13}}{6} \right) \][/tex]
After this, we identify the four specified values in the provided solution which are:
[tex]\[ x = -\tan^{-1} \left( -\frac{1}{2} + \frac{\sqrt{2}\sqrt{5 - \sqrt{13}}}{2} + \frac{\sqrt{13}}{2} \right) \][/tex]
[tex]\[ x = \tan^{-1} \left( \frac{1}{2} + \frac{\sqrt{13}}{2} + \frac{\sqrt{2\sqrt{13} + 10}}{2} \right) \][/tex]
[tex]\[ x = \tan^{-1} \left( -\frac{\sqrt{13}}{2} + \frac{1}{2} + \frac{\sqrt{10 - 2\sqrt{13}}}{2} \right) \][/tex]
[tex]\[ x = \tan^{-1} \left( -\frac{\sqrt{2\sqrt{13} + 10}}{2} + \frac{1}{2} + \frac{\sqrt{13}}{2} \right) \][/tex]
Those values fit perfectly within the given range [tex]\[-\frac{\pi}{2}, \frac{\pi}{2}\][/tex].
