6.3 If $$ \quad an x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} $$ and $$ m^{2}+\frac{1}{m^{2}}=1 $$ Calculate the value of $$ x $$ without the use of a calculator.

Question

6.3 If $$ \quad 	an x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} $$ and $$ m^{2}+\frac{1}{m^{2}}=1 $$ Calculate the value of $$ x $$ without the use of a calculator.

UpStudy AI · Accepted Answer

To solve the problem, we start with the given equations:

1. $$ \tan x = m + \frac{1}{m} $$
2. $$ m^2 + \frac{1}{m^2} = 1 $$

### Step 1: Solve for $$ m $$

From the second equation, we can express $$ m^2 + \frac{1}{m^2} $$ in terms of $$ m $$:

$$
m^2 + \frac{1}{m^2} = \left(m + \frac{1}{m}\right)^2 - 2
$$

Let $$ y = m + \frac{1}{m} $$. Then we have:

$$
m^2 + \frac{1}{m^2} = y^2 - 2
$$

Setting this equal to 1 gives:

$$
y^2 - 2 = 1
$$

Solving for $$ y $$:

$$
y^2 = 3 \implies y = \sqrt{3} \text{ or } y = -\sqrt{3}
$$

### Step 2: Determine $$ m $$

Now we have two cases for $$ y $$:

1. $$ m + \frac{1}{m} = \sqrt{3} $$
2. $$ m + \frac{1}{m} = -\sqrt{3} $$

#### Case 1: $$ m + \frac{1}{m} = \sqrt{3} $$

Multiplying both sides by $$ m $$:

$$
m^2 - \sqrt{3}m + 1 = 0
$$

Using the quadratic formula:

$$
m = \frac{\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \cdot 1}}{2} = \frac{\sqrt{3} \pm \sqrt{-1}}{2}
$$

This gives us complex solutions, which we will not consider further.

#### Case 2: $$ m + \frac{1}{m} = -\sqrt{3} $$

Similarly, we multiply both sides by $$ m $$:

$$
m^2 + \sqrt{3}m + 1 = 0
$$

Using the quadratic formula again:

$$
m = \frac{-\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \cdot 1}}{2} = \frac{-\sqrt{3} \pm \sqrt{-1}}{2}
$$

Again, we get complex solutions.

### Step 3: Evaluate $$ \tan x $$

Since both cases lead to complex solutions for $$ m $$, we can evaluate $$ \tan x $$ directly from the original equation:

1. If $$ m + \frac{1}{m} = \sqrt{3} $$, then $$ \tan x = \sqrt{3} $$.
2. If $$ m + \frac{1}{m} = -\sqrt{3} $$, then $$ \tan x = -\sqrt{3} $$.

### Step 4: Find $$ x $$

Now we need to find $$ x $$ in the range $$ 90^{\circ} \leq x \leq 270^{\circ} $$:

- For $$ \tan x = \sqrt{3} $$:
  - The angle $$ x $$ that satisfies this in the specified range is $$ 120^{\circ} $$ (since $$ \tan 120^{\circ} = -\sqrt{3} $$).
  
- For $$ \tan x = -\sqrt{3} $$:
  - The angle $$ x $$ that satisfies this in the specified range is $$ 240^{\circ} $$ (since $$ \tan 240^{\circ} = \sqrt{3} $$).

### Conclusion

Thus, the possible values of $$ x $$ are:

$$
x = 120^{\circ} \quad \text{or} \quad x = 240^{\circ}
$$

However, since we are looking for $$ \tan x = -\sqrt{3} $$, the final answer is:

$$
\boxed{240^{\circ}}
$$
$$ x = 240^{\circ} $$

6.3 If and
Calculate the value of without the use of a calculator.

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Trigonometry Questions

6.3 If and Calculate the value of without the use of a calculator.