[Solved]: sin2x+cosx=2

To solve the equation:

\[
\sin(2x) + \cos(x) = 2
\]

we can analyze the range of the trigonometric functions involved.

### Step 1: Understand the Maximum Values

- **\(\sin(2x)\):** The sine function oscillates between \(-1\) and \(1\). Therefore, \(\sin(2x) \leq 1\).
- **\(\cos(x)\):** Similarly, the cosine function also oscillates between \(-1\) and \(1\). Thus, \(\cos(x) \leq 1\).

Adding these two functions:

\[
\sin(2x) + \cos(x) \leq 1 + 1 = 2
\]

The equation \(\sin(2x) + \cos(x) = 2\) implies that both \(\sin(2x)\) and \(\cos(x)\) must simultaneously reach their maximum values of \(1\).

### Step 2: Find When Each Function Reaches 1

1. **\(\sin(2x) = 1\)**
   \[
   2x = \frac{\pi}{2} + 2\pi n \quad \Rightarrow \quad x = \frac{\pi}{4} + \pi n \quad \text{where } n \text{ is an integer}
   \]

2. **\(\cos(x) = 1\)**
   \[
   x = 2\pi k \quad \text{where } k \text{ is an integer}
   \]

### Step 3: Check for Common Solutions

For both conditions to be satisfied simultaneously:

\[
\frac{\pi}{4} + \pi n = 2\pi k
\]

Rearranging:

\[
\frac{\pi}{4} = 2\pi k - \pi n \quad \Rightarrow \quad \frac{1}{4} = 2k - n
\]

Here, \(2k - n\) must be an integer since both \(k\) and \(n\) are integers. However, \(\frac{1}{4}\) is not an integer, which leads to a contradiction.

### Conclusion

There are **no real solutions** to the equation \(\sin(2x) + \cos(x) = 2\) because it's impossible for both \(\sin(2x)\) and \(\cos(x)\) to simultaneously reach their maximum value of \(1\) in a way that satisfies the equation.

**Final Answer:**  
There are no real numbers x that satisfy the equation sin 2x + cosx = 2.

Home Question Bank Math Trigonometry

Question

Ask by Hart Hill. in South Africa

Dec 21,2024

Upstudy AI Solution

Tutor-Verified Answer

Answer

There are no real solutions to the equation

Solution

To solve the equation:

we can analyze the range of the trigonometric functions involved.

Step 1: Understand the Maximum Values

: The sine function oscillates between and . Therefore, .
: Similarly, the cosine function also oscillates between and . Thus, .

Adding these two functions:

The equation

implies that both

and

must simultaneously reach their maximum values of

Step 2: Find When Each Function Reaches 1

Step 3: Check for Common Solutions

For both conditions to be satisfied simultaneously:

Rearranging:

Here,

must be an integer since both

and

are integers. However,

is not an integer, which leads to a contradiction.

Conclusion

There are no real solutions to the equation

because it’s impossible for both

and

to simultaneously reach their maximum value of

in a way that satisfies the equation.

Final Answer:
There are no real numbers x that satisfy the equation sin 2x + cosx = 2.

Answered by UpStudy AI and reviewed by a Professional Tutor

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Explain

Simplify this solution

Extra Insights

To solve the equation

, it’s important to note that both

and

range between -1 and 1 for all real values of

. The maximum value of

occurs when both functions equal 1. Therefore, this means

must also equal 1—which happens when

for any integer

—while simultaneously,

must equal 1, which occurs when

However, you can quickly see that

can only reach 1 while

is also equal to 1 under specific conditions. Due to these constraints, the only solution to this equation is when both equal 1 simultaneously, leading to no real solutions. This elegant dance between sine and cosine demonstrates the intricate beauty of trigonometry!

When grappling with trigonometric equations like this, one common pitfall is overlooking the fact that functions like sine and cosine oscillate between fixed values. It’s easy to misinterpret whether a sum can actually reach its theoretical maximum, so actively checking value ranges can prevent unnecessary calculation errors. Always keep in mind the periodic nature of these trigonometric functions!

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