Question

2Sin(t) + 2Cos(t) = 1.5

Answer 1

To solve the equation \( 2\sin(t) + 2\cos(t) = 1.5 \), we can start by simplifying it.

1. Divide both sides by 2:

\[
\sin(t) + \cos(t) = \frac{1.5}{2} = 0.75
\]

So, we rewrite the equation:

\[
\sin(t) + \cos(t) = 0.75
\]

2. Next, we can square both sides to eliminate the sine and cosine:

\[
(\sin(t) + \cos(t))^2 = (0.75)^2
\]

Expanding the left-hand side, we have:

\[
\sin^2(t) + 2\sin(t)\cos(t) + \cos^2(t) = 0.5625
\]

Using the Pythagorean identity \(\sin^2(t) + \cos^2(t) = 1\), the equation simplifies to:

\[
1 + 2\sin(t)\cos(t) = 0.5625
\]

3. Rearranging gives:

\[
2\sin(t)\cos(t) = 0.5625 - 1 = -0.4375
\]

4. We know that \(2\sin(t)\cos(t) = \sin(2t)\). Thus, we rewrite the equation as:

\[
\sin(2t) = -0.4375
\]

5. To solve for \(2t\), we can take the inverse sine:

\[
2t = \arcsin(-0.4375)
\]

This gives two solutions for \(2t\) in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\):

\[
2t = -0.4524 \quad \text{or} \quad 2t = \pi - (-0.4524) \approx 3.5930
\]

6. Now, we solve for \(t\):

1. From \(2t = -0.4524\):
\[
t = -0.2262
\]

2. From \(2t \approx 3.5930\):
\[
t \approx 1.7965
\]

7. Since the sine function is periodic, we can add multiples of \( \pi \) to these solutions to get the general solutions:

\[
t = -0.2262 + n\pi \quad \text{for any integer } n
\]
\[
t \approx 1.7965 + n\pi \quad \text{for any integer } n
\]

These give us the complete set of solutions for the original equation \( 2\sin(t) + 2\cos(t) = 1.5 \).