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Math  /  Trigonometry

Question5sin⁡xcos⁡x+cos⁡x=05 \sin x \cos x+\cos x=0

Studdy Solution

STEP 1

1. We are solving the trigonometric equation 5sin⁡xcos⁡x+cos⁡x=05 \sin x \cos x + \cos x = 0.
2. The goal is to find the values of xx that satisfy this equation.
3. We will use trigonometric identities and algebraic manipulation to solve the equation.

STEP 2

1. Factor the equation.
2. Solve each factor for xx.
3. Determine the general solutions.
4. Consider the domain of the solutions if necessary.

STEP 3

Factor the equation 5sin⁡xcos⁡x+cos⁡x=05 \sin x \cos x + \cos x = 0.
Notice that cos⁡x\cos x is a common factor:
cos⁡x(5sin⁡x+1)=0 \cos x (5 \sin x + 1) = 0

STEP 4

Solve each factor for xx.
First, solve cos⁡x=0\cos x = 0:
cos⁡x=0\cos x = 0
The solutions for cos⁡x=0\cos x = 0 are:
x=π2+nπ,n∈Zx = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}

STEP 5

Next, solve 5sin⁡x+1=05 \sin x + 1 = 0:
5sin⁡x+1=05 \sin x + 1 = 0
Subtract 1 from both sides:
5sin⁡x=−15 \sin x = -1
Divide by 5:
sin⁡x=−15\sin x = -\frac{1}{5}
The solutions for sin⁡x=−15\sin x = -\frac{1}{5} are:
x=arcsin⁡(−15)+2kπorx=π−arcsin⁡(−15)+2kπ,k∈Zx = \arcsin\left(-\frac{1}{5}\right) + 2k\pi \quad \text{or} \quad x = \pi - \arcsin\left(-\frac{1}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}

STEP 6

Combine the solutions from both factors to express the general solution:
The general solutions are:
1. x=π2+nπ,n∈Zx = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}
2. x=arcsin⁡(−15)+2kπ,k∈Zx = \arcsin\left(-\frac{1}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}
3. x=π−arcsin⁡(−15)+2kπ,k∈Zx = \pi - \arcsin\left(-\frac{1}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}

STEP 7

Consider the domain of the solutions if necessary. If no specific domain is given, the solutions are valid for all xx.
The solutions are:
1. x=π2+nπ,n∈Zx = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}
2. x=arcsin⁡(−15)+2kπ,k∈Zx = \arcsin\left(-\frac{1}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}
3. x=π−arcsin⁡(−15)+2kπ,k∈Zx = \pi - \arcsin\left(-\frac{1}{5}\right) + 2k\pi, \quad k \in \mathbb{Z}

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