Let's solve the given trigonometric equation step-by-step:
### Given Equation:
[tex]\[ 7 \sin(t) + 5 = 2 \sin(t) \][/tex]
### Part (a): Solving for all radian solutions
1. Isolate [tex]\(\sin(t)\)[/tex]:
To solve for [tex]\(\sin(t)\)[/tex], we need to first isolate the term [tex]\(\sin(t)\)[/tex] on one side of the equation. So we start by subtracting [tex]\(2 \sin(t)\)[/tex] from both sides of the equation:
[tex]\[
7 \sin(t) + 5 - 2 \sin(t) = 2 \sin(t) - 2 \sin(t)
\][/tex]
Simplifying this, we get:
[tex]\[
5 \sin(t) + 5 = 0
\][/tex]
2. Solve for [tex]\(\sin(t)\)[/tex]:
Next, we move the constant term to the other side of the equation by subtracting 5 from both sides:
[tex]\[
5 \sin(t) = -5
\][/tex]
Now, divide both sides by 5 to solve for [tex]\(\sin(t)\)[/tex]:
[tex]\[
\sin(t) = -1
\][/tex]
3. Determine all radian solutions:
The standard equation for [tex]\(\sin(t) = -1\)[/tex] is satisfied at specific values of [tex]\(t\)[/tex]. In one full period of the sine function, [tex]\(\sin(t) = -1\)[/tex] occurs at:
[tex]\[
t = \frac{3\pi}{2}
\][/tex]
Since sine has a period of [tex]\(2\pi\)[/tex], the general solution for all radian values [tex]\(t\)[/tex] where [tex]\(\sin(t) = -1\)[/tex] is:
[tex]\[
t = \frac{3\pi}{2} + 2k\pi
\][/tex]
where [tex]\(k\)[/tex] is any integer.
### Part (b): Solving within the range [tex]\(0 \leq t < 2\pi\)[/tex]
To find the specific solution within one complete cycle of the sine function, [tex]\(0 \leq t < 2\pi\)[/tex], we consider the standard interval:
- Within [tex]\(0 \leq t < 2\pi\)[/tex], the value [tex]\(\sin(t) = -1\)[/tex] occurs at:
[tex]\[
t = \frac{3\pi}{2}
\][/tex]
Thus, the solution to the equation [tex]\(\sin(t) = -1\)[/tex] within the specified range is:
[tex]\[
t = \frac{3\pi}{2}
\][/tex]
### Final Answer
- Part (a): The general solution in radians is:
[tex]\[
t = \frac{3\pi}{2} + 2k\pi
\][/tex]
where [tex]\(k\)[/tex] is any integer.
- Part (b): The specific solution within [tex]\(0 \leq t < 2\pi\)[/tex] is:
[tex]\[
t = \frac{3\pi}{2}
\][/tex]
