"ror the qucsmons that hallow, you may asvume that $$ x_{p}=30^{\circ} $$ and $$ \xi_{Q}=150^{\circ} $$ (4) For math talues of $$ x $$, in the interyal $$ \left|0^{0}, 180^{\circ}\right| $$, is (i) $$ f(x) \leq 07 $$ (ii) $$ f(x)g(x) $$ ? (5) Lis the given graphe, and your calculations above, to determine the valuc(s) of $$ x $$, in the interval $$ \left(0^{2} ; 180^{2}\right) $$. for which (i) $$ \sin ^{2} x-\cos ^{2} x=1 $$ (ii) $$ 2 \cos ^{2} x=3 \sin x $$ (6) Esplain howt the graph of $$ f $$ can be tramoformod in rovetune the graph of (i) $$ x(x)=2 \cos ^{2} x $$ (B) $$ \quad(x)=2 \sin ^{2} x-1 $$

Question

"ror the qucsmons that hallow, you may asvume that $$ x_{p}=30^{\circ} $$ and $$ \xi_{Q}=150^{\circ} $$ (4) For math talues of $$ x $$, in the interyal $$ \left|0^{0}, 180^{\circ}\right| $$, is (i) $$ f(x) \leq 07 $$ (ii) $$ f(x)>g(x) $$ ? (5) Lis the given graphe, and your calculations above, to determine the valuc(s) of $$ x $$, in the interval $$ \left(0^{2} ; 180^{2}\right) $$. for which (i) $$ \sin ^{2} x-\cos ^{2} x=1 $$ (ii) $$ 2 \cos ^{2} x=3 \sin x $$ (6) Esplain howt the graph of $$ f $$ can be tramoformod in rovetune the graph of (i) $$ x(x)=2 \cos ^{2} x $$ (B) $$ \quad(x)=2 \sin ^{2} x-1 $$

UpStudy AI · Accepted Answer

Let's break down the problem step by step, extracting the known conditions and solving each part systematically.

### Known Conditions
1. $$ x_{p} = 30^{\circ} $$
2. $$ \xi_{Q} = 150^{\circ} $$
3. We need to analyze the functions $$ f(x) $$ and $$ g(x) $$ in the interval $$ (0^{\circ}, 180^{\circ}) $$.
4. We have two equations to solve:
   - $$ \sin^2 x - \cos^2 x = 1 $$
   - $$ 2 \cos^2 x = 3 \sin x $$

### Part (4)
#### (i) Solve $$ f(x) \leq 07 $$
Since we don't have the explicit form of $$ f(x) $$ or $$ g(x) $$, we cannot solve this part directly. However, we can assume that $$ f(x) $$ is a trigonometric function based on the context.

#### (ii) Solve $$ f(x) > g(x) $$
Again, without the explicit forms of $$ f(x) $$ and $$ g(x) $$, we cannot solve this part directly.

### Part (5)
Now, let's solve the equations given in this part.

#### (i) Solve $$ \sin^2 x - \cos^2 x = 1 $$
We can rewrite this equation using the identity $$ \sin^2 x + \cos^2 x = 1 $$:
$$
\sin^2 x - (1 - \sin^2 x) = 1
$$
This simplifies to:
$$
2\sin^2 x - 1 = 1
$$
$$
2\sin^2 x = 2
$$
$$
\sin^2 x = 1
$$
Thus, $$ \sin x = 1 $$ or $$ \sin x = -1 $$. In the interval $$ (0^{\circ}, 180^{\circ}) $$, the only solution is:
$$
x = 90^{\circ}
$$

#### (ii) Solve $$ 2 \cos^2 x = 3 \sin x $$
We can rewrite this equation as:
$$
2(1 - \sin^2 x) = 3 \sin x
$$
This leads to:
$$
2 - 2\sin^2 x = 3\sin x
$$
Rearranging gives us a quadratic equation:
$$
2\sin^2 x + 3\sin x - 2 = 0
$$
Let $$ y = \sin x $$. The equation becomes:
$$
2y^2 + 3y - 2 = 0
$$
We can solve this quadratic equation using the quadratic formula:
$$
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$ a = 2, b = 3, c = -2 $$.

Calculating the discriminant:
$$
b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25
$$
Now substituting into the quadratic formula:
$$
y = \frac{-3 \pm 5}{4}
$$
This gives us two solutions:
1. $$ y = \frac{2}{4} = \frac{1}{2} $$
2. $$ y = \frac{-8}{4} = -2 $$ (not valid since $$ \sin x $$ must be between -1 and 1)

Thus, we have:
$$
\sin x = \frac{1}{2}
$$
In the interval $$ (0^{\circ}, 180^{\circ}) $$, this corresponds to:
$$
x = 30^{\circ}
$$

### Summary of Solutions
1. For $$ \sin^2 x - \cos^2 x = 1 $$, the solution is $$ x = 90^{\circ} $$.
2. For $$ 2 \cos^2 x = 3 \sin x $$, the solution is $$ x = 30^{\circ} $$.

### Part (6)
#### (i) Transforming $$ f(x) = 2 \cos^2 x $$
The function $$ f(x) = 2 \cos^2 x $$ can be transformed into $$ g(x) = 2 \sin^2 x - 1 $$ using the identity $$ \cos^2 x = 1 - \sin^2 x $$:
$$
f(x) = 2(1 - \sin^2 x) = 2 - 2\sin^2 x
$$
To transform $$ f(x) $$ into $$ g(x) $$, we can express it as:
$$
g(x) = 2\sin^2 x - 1
$$
This shows how the graph of $$ f(x) $$ can be transformed to obtain $$ g(x) $$.

### Conclusion
The solutions for the equations are $$ x = 90^{\circ} $$ and $$ x = 30^{\circ} $$. The transformation of the graphs can be explained using trigonometric identities.
The solutions are: 1. $$ x = 90^{\circ} $$ for $$ \sin^2 x - \cos^2 x = 1 $$ 2. $$ x = 30^{\circ} $$ for $$ 2 \cos^2 x = 3 \sin x $$ The graph of $$ f(x) = 2 \cos^2 x $$ can be transformed into $$ g(x) = 2 \sin^2 x - 1 $$ using trigonometric identities.

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