Math  /  Trigonometry

Question2sinxcosx=3sinx2 \sin x \cos x=3 \sin x

Studdy Solution

STEP 1

What is this asking? We're looking for all the angles, xx, that make this *trigonometric equation* true! Watch out! Don't divide both sides by sin(x)\sin(x)!
You might lose some solutions!

STEP 2

1. Rewrite the equation
2. Consider the zero case
3. Consider the non-zero case

STEP 3

Alright, let's **rewrite** our equation to make it easier to work with!
We'll move everything to one side by subtracting 3sin(x)3\sin(x) from both sides.
This gives us: 2sin(x)cos(x)3sin(x)=02\sin(x)\cos(x) - 3\sin(x) = 0

STEP 4

Now, we can **factor out** sin(x)\sin(x), which is super helpful!
This gives us: sin(x)(2cos(x)3)=0\sin(x) \cdot (2\cos(x) - 3) = 0

STEP 5

This equation is true if either sin(x)=0\sin(x) = 0 or 2cos(x)3=02\cos(x) - 3 = 0.
Let's **tackle** sin(x)=0\sin(x) = 0 first.
When does sin(x)\sin(x) equal **zero**?

STEP 6

Well, sin(x)=0\sin(x) = 0 when xx is a **multiple** of π\pi!
So, x=nπx = n\pi, where nn can be any **integer** (like 0, 1, 2, -1, -2, and so on).
We've found some solutions already!

STEP 7

Now, let's **explore** the other possibility: 2cos(x)3=02\cos(x) - 3 = 0.
First, we **isolate** cos(x)\cos(x) by adding 3 to both sides and then dividing by 2: 2cos(x)=32\cos(x) = 3 cos(x)=32\cos(x) = \frac{3}{2}

STEP 8

Hold on a second!
The **cosine function** can never be greater than **one** or less than **negative one**.
Since 32\frac{3}{2} is greater than one, there are no solutions for this part of the equation!
That means we've already found all the solutions in the previous step!

STEP 9

The solutions to the equation 2sin(x)cos(x)=3sin(x)2\sin(x)\cos(x) = 3\sin(x) are x=nπx = n\pi, where nn is any integer.

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