Finding tanx from Trigonometric Equation

We are given a trigonometric equation involving $\sin x$ and $\cos x$, and we need to find the value of $\tan x$. The given equation is:

$\sin x + \cos x = \sqrt{2}\sin x$

Solving the Trigonometric Equation for tanx

Our goal is to isolate $\tan x$, which is defined as $\frac{\sin x}{\cos x}$. To achieve this, we need to manipulate the given equation to get terms involving $\sin x$ and $\cos x$ on opposite sides.

Step-by-Step Solution

Let's solve the equation step by step:

1. Start with the given equation:

   $\sin x + \cos x = \sqrt{2}\sin x$

2. Rearrange the terms to group $\sin x$ terms together. Subtract $\sin x$ from both sides of the equation:

   $\cos x = \sqrt{2}\sin x - \sin x$

3. Factor out $\sin x$ from the terms on the right-hand side:

   $\cos x = (\sqrt{2} - 1)\sin x$

4. Now, to get $\tan x = \frac{\sin x}{\cos x}$, we need to divide both sides by $\cos x$. Before dividing, consider if $\cos x$ can be zero. If $\cos x = 0$, then the original equation becomes $\sin x = \sqrt{2}\sin x$. This implies $(\sqrt{2}-1)\sin x = 0$. Since $\sqrt{2}-1 \neq 0$, we must have $\sin x = 0$. However, $\sin x$ and $\cos x$ cannot both be zero for the same angle $x$ (as $\sin^2 x + \cos^2 x = 1$). Therefore, $\cos x$ cannot be zero, and we can safely divide by $\cos x$:

   $\frac{\cos x}{\cos x} = \frac{(\sqrt{2} - 1)\sin x}{\cos x}$

5. Simplify both sides. On the left, $\frac{\cos x}{\cos x} = 1$. On the right, $\frac{\sin x}{\cos x} = \tan x$.

   $1 = (\sqrt{2} - 1)\tan x$

6. Solve for $\tan x$ by dividing both sides by $(\sqrt{2} - 1)$:

   $\tan x = \frac{1}{\sqrt{2} - 1}$

7. The denominator contains a radical, so we should rationalize it. Multiply the numerator and denominator by the conjugate of the denominator, which is $(\sqrt{2} + 1)$:

   $\tan x = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$

8. Use the difference of squares formula in the denominator: $(a-b)(a+b) = a^2 - b^2$. Here, $a = \sqrt{2}$ and $b = 1$.

   $\tan x = \frac{\sqrt{2} + 1}{(\sqrt{2})^2 - (1)^2}$

9. Simplify the denominator:

   $\tan x = \frac{\sqrt{2} + 1}{2 - 1}$

   $\tan x = \frac{\sqrt{2} + 1}{1}$

10. Final value of $\tan x$:

    $\tan x = \sqrt{2} + 1$

Summary of the Solution Steps

We started with the given equation $\sin x + \cos x = \sqrt{2}\sin x$. By moving terms and factoring, we obtained $\cos x = (\sqrt{2} - 1)\sin x$. Dividing by $\cos x$ (which is non-zero), we got $1 = (\sqrt{2} - 1)\tan x$. Finally, solving for $\tan x$ and rationalizing the denominator gave us $\tan x = \sqrt{2} + 1$.

Revision Table: Key Trigonometry Concepts

Additional Information: Solving Trigonometric Equations

Solving trigonometric equations often involves using identities to simplify expressions, rearranging terms, factoring, and isolating the trigonometric function. Here are some general strategies:

• If the equation involves different trigonometric functions (like $\sin x$ and $\cos x$), try to express them in terms of a single function, or relate them using identities like $\tan x = \frac{\sin x}{\cos x}$.
• Look for opportunities to factor expressions.
• If the equation is quadratic in form (e.g., involving $\sin^2 x$), consider using the Pythagorean identity ($\sin^2 x + \cos^2 x = 1$) to express everything in terms of one function, and then solve the quadratic.
• Be mindful of the domain and range of trigonometric functions and potential restrictions when dividing or taking square roots.
• When solving for the angle $x$ itself (not required in this problem), remember that trigonometric functions are periodic and may have multiple solutions.

This problem focused on algebraic manipulation of a trigonometric equation to find the value of a specific trigonometric ratio, $\tan x$.