Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

Solve the equation

[tex]\[ \cos \left(\theta+20^{\circ}\right) - \cos \left(\theta+38^{\circ}\right) = 0.3 \][/tex]

for [tex]\[ 0^{\circ} \leq \theta \leq 360^{\circ} \][/tex]


Sagot :

To solve the equation [tex]\(\cos \left(\theta + 20^{\circ}\right) - \cos \left(\theta + 38^{\circ}\right) = 0.3\)[/tex] for [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex], we need to go through the following steps:

1. Rewrite the equation using trigonometric identities if necessary:
Given equation:
[tex]\[ \cos \left(\theta + 20^{\circ}\right) - \cos \left(\theta + 38^{\circ}\right) = 0.3 \][/tex]

2. Express cosine differences in a simplified form:
While there are trigonometric identities for certain transformations of cosine differences, the given difference is not straightforward to simplify cleanly using standard identities. However, we directly work with it in its given form.

3. Numerical solution approach:
To find the exact solutions, we typically solve the equation numerically within the specified interval.

4. Solving for [tex]\(\theta\)[/tex] in the range [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex]:
By analyzing the equation [tex]\(\cos \left(\theta + 20^{\circ}\right) - \cos \left(\theta + 38^{\circ}\right) = 0.3\)[/tex] numerically within the interval [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex], we find that there are no values of [tex]\(\theta\)[/tex] that satisfy the equation within this interval.

Therefore, the solution set for the given equation in the specified range of [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex] is:

[tex]\[ \boxed{[]} \][/tex]
This indicates that there are no solutions to the equation within the given interval.