Sure! Let's solve the equation \((\tan (\theta)-2)\left(25 \sin^2(\theta)-1\right)=0\).
### Step-by-Step Solution:
This equation is a product of two factors, and for the product to be zero, at least one of the factors must be zero. Therefore, we'll break it down into two separate equations and solve them individually.
#### Equation 1: \(\tan(\theta) - 2 = 0\)
1. Isolate \(\tan(\theta)\):
[tex]\[
\tan(\theta) - 2 = 0
\][/tex]
[tex]\[
\tan(\theta) = 2
\][/tex]
2. Solve for \(\theta\):
We need to find \(\theta\) such that the tangent of \(\theta\) equals 2. One solution is:
[tex]\[
\theta = \tan^{-1}(2)
\][/tex]
This can be written as:
[tex]\[
\theta = \arctan(2)
\][/tex]
#### Equation 2: \(25 \sin^2(\theta) - 1 = 0\)
1. Isolate \(\sin^2(\theta)\):
[tex]\[
25 \sin^2(\theta) - 1 = 0
\][/tex]
[tex]\[
25 \sin^2(\theta) = 1
\][/tex]
[tex]\[
\sin^2(\theta) = \frac{1}{25}
\][/tex]
2. Solve for \(\sin(\theta)\):
We need to find \(\theta\) such that the sine of \(\theta\) squared equals \(\frac{1}{25}\). This gives:
[tex]\[
\sin(\theta) = \pm \frac{1}{5}
\][/tex]
So, there are two possible solutions:
[tex]\[
\sin(\theta) = \frac{1}{5}
\][/tex]
[tex]\[
\sin(\theta) = -\frac{1}{5}
\][/tex]
3. Solve for \(\theta\):
[tex]\[
\theta = \sin^{-1}\left(\frac{1}{5}\right)
\][/tex]
[tex]\[
\theta = \sin^{-1}\left(-\frac{1}{5}\right)
\][/tex]
These can be written more specifically as:
[tex]\[
\theta = \arcsin\left(\frac{1}{5}\right) \approx 0.201357920790331
\][/tex]
[tex]\[
\theta = \arcsin\left(-\frac{1}{5}\right) \approx -0.201357920790331
\][/tex]
### Summary of Solutions:
The solutions to the equation \((\tan (\theta)-2)\left(25 \sin^2(\theta)-1\right)=0\) are:
[tex]\[
\theta = \arctan(2)
\][/tex]
[tex]\[
\theta \approx 0.201357920790331
\][/tex]
[tex]\[
\theta \approx -0.201357920790331
\][/tex]
These represent the values of [tex]\(\theta\)[/tex] where either [tex]\(\tan(\theta) = 2\)[/tex] or [tex]\(\sin^2(\theta) = \frac{1}{25}\)[/tex].
