\sqrt{2}\cos \left(x-\frac{\pi }{4}\right)+\left(\sqrt{3}-1\right)\sin x=2

solve this equation for all possible values in radians within the interval of 0 to 2pi

To solve the equation
\[
\sqrt{2} \cos \left(x - \frac{\pi}{4}\right) + \left(\sqrt{3} - 1\right) \sin x = 2
\]
for \( x \) in the interval \( [0, 2\pi] \), let's proceed as follows:

1. **Expand the Cosine Term**:
   We can expand \( \cos \left(x - \frac{\pi}{4}\right) \) using the cosine difference identity:
   \[
   \cos \left(x - \frac{\pi}{4}\right) = \cos x \cos \frac{\pi}{4} + \sin x \sin \frac{\pi}{4}
   \]
   Since \( \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), we have:
   \[
   \cos \left(x - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x
   \]

2. **Substitute and Simplify**:
   Substitute this into the equation:
   \[
   \sqrt{2} \left(\frac{\sqrt{2}}{2} \cos x + \frac{\sqrt{2}}{2} \sin x\right) + \left(\sqrt{3} - 1\right) \sin x = 2
   \]
   Simplify the first term:
   \[
   \sqrt{2} \cdot \frac{\sqrt{2}}{2} \cos x + \sqrt{2} \cdot \frac{\sqrt{2}}{2} \sin x = \cos x + \sin x
   \]
   The equation becomes:
   \[
   \cos x + \sin x + \left(\sqrt{3} - 1\right) \sin x = 2
   \]
   Combine the sine terms:
   \[
   \cos x + \left(1 + \sqrt{3} - 1\right) \sin x = 2
   \]
   Simplify further:
   \[
   \cos x + \sqrt{3} \sin x = 2
   \]

3. **Express in Terms of a Single Sine or Cosine Function**:
   Rewrite \( \cos x + \sqrt{3} \sin x \) as \( R \cos(x - \alpha) \) for some \( R \) and \( \alpha \), where:
   \[
   R = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2
   \]
   and
   \[
   \cos \alpha = \frac{1}{2}, \quad \sin \alpha = \frac{\sqrt{3}}{2} \Rightarrow \alpha = \frac{\pi}{3}
   \]
   Thus,
   \[
   \cos x + \sqrt{3} \sin x = 2 \cos \left( x - \frac{\pi}{3} \right)
   \]
   Our equation now simplifies to:
   \[
   2 \cos \left( x - \frac{\pi}{3} \right) = 2
   \]
   Divide both sides by 2:
   \[
   \cos \left( x - \frac{\pi}{3} \right) = 1
   \]

4. **Solve for \( x \)**:
   The equation \( \cos \left( x - \frac{\pi}{3} \right) = 1 \) holds when
   \[
   x - \frac{\pi}{3} = 2k\pi \quad \text{for integers } k
   \]
   Solving for \( x \):
   \[
   x = \frac{\pi}{3} + 2k\pi
   \]
   Within the interval \( [0, 2\pi] \), this solution only holds for \( k = 0 \), so
   \[
   x = \frac{\pi}{3}
   \]

### Solution
The solution to the equation within \( [0, 2\pi] \) is:
\[
x = \frac{\pi}{3}
\]

Would you like more details on any step? 

---

Here are some related questions you might find useful:
1. How do trigonometric identities help in solving equations like this?
2. What does it mean to express an equation in terms of a single cosine or sine function?
3. Why does \( \cos(x - \frac{\pi}{3}) = 1 \) only have certain solutions?
4. How do we determine the interval for solutions in trigonometric equations?
5. How do we handle more complex trigonometric equations involving multiple terms?

**Tip:** For equations like this, simplifying expressions with identities can make them far easier to solve directly.

\sqrt{2}\cos \left(x-\frac{\pi }{4}\right)+\left(\sqrt{3}-1\right)\sin x=2. Solve this equation for all possible values in radians within the interval of 0 to 2pi.

This guide walks through solving the trigonometric equation \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) + \left(\sqrt{3} - 1\right) \sin x = 2 over the interval [0, 2π]. Learn to use cosine difference identities, transformation techniques, and trigonometric solving methods for precise solutions. Ideal for high school students in Grades 11-12.

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