Answer :
Sure, let's solve the equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] step-by-step:
1. Identify the type of equation: The equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] is a trigonometric equation.
2. Combine the sine and cosine terms: We can combine the sine and cosine terms by expressing them as a single sine function. This can be done by recognizing that [tex]\(\sin x - \sqrt{3} \cos x\)[/tex] can be written in the form [tex]\(R \sin(x + \phi)\)[/tex].
We know the identity:
[tex]\[ a \sin x + b \cos x = R \sin(x + \phi) \][/tex]
where [tex]\(R = \sqrt{a^2 + b^2}\)[/tex] and [tex]\(\tan \phi = \frac{b}{a}\)[/tex].
For our equation, [tex]\(a = 1\)[/tex] and [tex]\(b = -\sqrt{3}\)[/tex]:
[tex]\[ R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \][/tex]
So, we have:
[tex]\[ \sin x - \sqrt{3} \cos x = 2 \sin(x + \phi) \][/tex]
3. Find [tex]\(\phi\)[/tex]: To find [tex]\(\phi\)[/tex], we use [tex]\(\tan \phi = \frac{-\sqrt{3}}{1} = -\sqrt{3}\)[/tex].
The angle [tex]\(\phi\)[/tex] for which [tex]\(\tan \phi = -\sqrt{3}\)[/tex] is [tex]\(\phi = -\pi/3\)[/tex] (since [tex]\(\tan(-\pi/3) = -\sqrt{3}\)[/tex]).
4. Rewrite the equation: Substitute [tex]\(R = 2\)[/tex] and [tex]\(\phi = -\pi/3\)[/tex] into the equation:
[tex]\[ \sin x - \sqrt{3} \cos x = 2 \sin(x + (-\pi/3)) = 2 \sin(x - \pi/3) \][/tex]
So, the original equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] is equivalent to:
[tex]\[ 2 \sin(x - \pi/3) = 2 \][/tex]
5. Simplify: Divide both sides by 2:
[tex]\[ \sin(x - \pi/3) = 1 \][/tex]
6. Solve for [tex]\(x\)[/tex]: We need to find the values of [tex]\(x\)[/tex] such that [tex]\(\sin(x - \pi/3) = 1\)[/tex].
The sine function equals 1 at [tex]\( \frac{\pi}{2} + 2k\pi \)[/tex] where [tex]\( k \)[/tex] is any integer:
[tex]\[ x - \pi/3 = \frac{\pi}{2} + 2k\pi \][/tex]
Therefore, solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\pi}{2} + \pi/3 + 2k\pi \][/tex]
Simplifying the term [tex]\(\frac{\pi}{2} + \pi/3\)[/tex]:
[tex]\[ \frac{\pi}{2} + \pi/3 = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \][/tex]
So, the general solution is:
[tex]\[ x = \frac{5\pi}{6} + 2k\pi \quad \text{where } k \text{ is any integer} \][/tex]
7. Specific solutions: For [tex]\(k = 0\)[/tex], we get:
[tex]\[ x = \frac{5\pi}{6} \][/tex]
Thus, one specific solution to the equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] is:
[tex]\[ x = \frac{5\pi}{6} \][/tex]
1. Identify the type of equation: The equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] is a trigonometric equation.
2. Combine the sine and cosine terms: We can combine the sine and cosine terms by expressing them as a single sine function. This can be done by recognizing that [tex]\(\sin x - \sqrt{3} \cos x\)[/tex] can be written in the form [tex]\(R \sin(x + \phi)\)[/tex].
We know the identity:
[tex]\[ a \sin x + b \cos x = R \sin(x + \phi) \][/tex]
where [tex]\(R = \sqrt{a^2 + b^2}\)[/tex] and [tex]\(\tan \phi = \frac{b}{a}\)[/tex].
For our equation, [tex]\(a = 1\)[/tex] and [tex]\(b = -\sqrt{3}\)[/tex]:
[tex]\[ R = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \][/tex]
So, we have:
[tex]\[ \sin x - \sqrt{3} \cos x = 2 \sin(x + \phi) \][/tex]
3. Find [tex]\(\phi\)[/tex]: To find [tex]\(\phi\)[/tex], we use [tex]\(\tan \phi = \frac{-\sqrt{3}}{1} = -\sqrt{3}\)[/tex].
The angle [tex]\(\phi\)[/tex] for which [tex]\(\tan \phi = -\sqrt{3}\)[/tex] is [tex]\(\phi = -\pi/3\)[/tex] (since [tex]\(\tan(-\pi/3) = -\sqrt{3}\)[/tex]).
4. Rewrite the equation: Substitute [tex]\(R = 2\)[/tex] and [tex]\(\phi = -\pi/3\)[/tex] into the equation:
[tex]\[ \sin x - \sqrt{3} \cos x = 2 \sin(x + (-\pi/3)) = 2 \sin(x - \pi/3) \][/tex]
So, the original equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] is equivalent to:
[tex]\[ 2 \sin(x - \pi/3) = 2 \][/tex]
5. Simplify: Divide both sides by 2:
[tex]\[ \sin(x - \pi/3) = 1 \][/tex]
6. Solve for [tex]\(x\)[/tex]: We need to find the values of [tex]\(x\)[/tex] such that [tex]\(\sin(x - \pi/3) = 1\)[/tex].
The sine function equals 1 at [tex]\( \frac{\pi}{2} + 2k\pi \)[/tex] where [tex]\( k \)[/tex] is any integer:
[tex]\[ x - \pi/3 = \frac{\pi}{2} + 2k\pi \][/tex]
Therefore, solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\pi}{2} + \pi/3 + 2k\pi \][/tex]
Simplifying the term [tex]\(\frac{\pi}{2} + \pi/3\)[/tex]:
[tex]\[ \frac{\pi}{2} + \pi/3 = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \][/tex]
So, the general solution is:
[tex]\[ x = \frac{5\pi}{6} + 2k\pi \quad \text{where } k \text{ is any integer} \][/tex]
7. Specific solutions: For [tex]\(k = 0\)[/tex], we get:
[tex]\[ x = \frac{5\pi}{6} \][/tex]
Thus, one specific solution to the equation [tex]\(\sin x - \sqrt{3} \cos x = 2\)[/tex] is:
[tex]\[ x = \frac{5\pi}{6} \][/tex]
