(i) If $$ 2-\cos ^{2} heta=3 \sin heta \cos heta $$, find the possible values of $$ an heta $$. (ii) If $$ \sin heta
eq \cos heta $$, explain how your answer in part (i) would be affected.

Question

(i) If $$ 2-\cos ^{2} 	heta=3 \sin 	heta \cos 	heta $$, find the possible values of $$ 	an 	heta $$. (ii) If $$ \sin 	heta 
eq \cos 	heta $$, explain how your answer in part (i) would be affected.

UpStudy AI · Accepted Answer

Given
$$
2-\cos^2\theta=3\sin\theta\cos\theta,
$$
we wish to solve for the possible values of $$\tan\theta$$.

**Step 1. Check for $$\cos\theta=0$$:**

If $$\cos\theta=0$$, then the left-hand side is
$$
2-\cos^2\theta=2-0=2,
$$
while the right-hand side is
$$
3\sin\theta\cos\theta=0.
$$
Since $$2 \neq 0$$, $$\cos\theta$$ cannot be zero.

**Step 2. Divide by $$\cos^2\theta$$:**

Because $$\cos\theta \neq 0$$, we can safely divide both sides of the equation by $$\cos^2\theta$$:
$$
\frac{2}{\cos^2\theta}-\frac{\cos^2\theta}{\cos^2\theta}=\frac{3\sin\theta\cos\theta}{\cos^2\theta}.
$$
This simplifies to
$$
\frac{2}{\cos^2\theta}-1=3\frac{\sin\theta}{\cos\theta}.
$$
Recognize that $$\frac{1}{\cos^2\theta}=\sec^2\theta=1+\tan^2\theta$$ and $$\frac{\sin\theta}{\cos\theta}=\tan\theta$$. Thus, the equation becomes
$$
2(1+\tan^2\theta)-1=3\tan\theta.
$$

**Step 3. Simplify the equation:**

Expanding and combining like terms:
$$
2+2\tan^2\theta-1=3\tan\theta,
$$
which simplifies to
$$
2\tan^2\theta+1=3\tan\theta.
$$
Rearrange this into standard quadratic form:
$$
2\tan^2\theta-3\tan\theta+1=0.
$$

**Step 4. Factor and solve the quadratic:**

Factor the quadratic:
$$
(2\tan\theta-1)(\tan\theta-1)=0.
$$
Thus, we have two possible cases:
$$
2\tan\theta-1=0\quad \text{or}\quad \tan\theta-1=0.
$$
Solving these:
- For $$2\tan\theta-1=0$$: $$\tan\theta=\frac{1}{2}$$.
- For $$\tan\theta-1=0$$: $$\tan\theta=1$$.

**Answer to (i):**

The possible values of $$\tan\theta$$ are $$\frac{1}{2}$$ and $$1$$.

**Step 5. Explain the effect of $$\sin\theta\neq\cos\theta$$:**

If $$\sin\theta\neq\cos\theta$$, note that $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$. For $$\tan\theta$$ to equal $$1$$, we must have
$$
\frac{\sin\theta}{\cos\theta}=1 \quad \Longrightarrow \quad \sin\theta=\cos\theta.
$$
Thus, if we are given that $$\sin\theta\neq\cos\theta$$, the solution $$\tan\theta=1$$ is excluded.

**Answer to (ii):**

When $$\sin\theta\neq\cos\theta$$, the value $$\tan\theta=1$$ is not allowed, leaving only $$\tan\theta=\frac{1}{2}$$ as the valid solution.
The possible values of $$\tan\theta$$ are $$\frac{1}{2}$$ and $$1$$. If $$\sin\theta \neq \cos\theta$$, then $$\tan\theta = 1$$ is not allowed, so the only valid solution is $$\tan\theta = \frac{1}{2}$$.

(i) If \( 2-\cos ^{2} \theta=3 \sin \theta \cos \theta \), find the possible values of \( \tan \theta \). (ii) If \( \sin \theta \neq \cos \theta \), explain how your answer in part (i) would be affected.

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