Question

solve the equation for t on the interval 0,2π).
2 sin t + 2 cos t = - 2
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a.
t = 0, - π/2
(simplify your answer. type your answer in radians. type an exact answer in terms of π. use integers or
fractions for any numbers in the expression. use a comma to separate answers as needed.)
b. there is no solution.

Explanation:

Step1: Divide both sides by 2

Divide the equation \(2\sin t + 2\cos t=-2\) by 2. We get \(\sin t+\cos t=-1\).

Step2: Square both sides

Square the equation \(\sin t+\cos t = - 1\). Using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), we have \((\sin t+\cos t)^{2}=(-1)^{2}\). So \(\sin^{2}t + 2\sin t\cos t+\cos^{2}t = 1\). Since \(\sin^{2}t+\cos^{2}t = 1\), the equation becomes \(1 + 2\sin t\cos t=1\), then \(2\sin t\cos t=0\), i.e., \(\sin(2t)=0\). So \(2t = k\pi\), \(t=\frac{k\pi}{2}\), \(k\in\mathbb{Z}\).

Step3: Check solutions in the original equation

• When \(t = 0\): \(2\sin(0)+2\cos(0)=2\times0 + 2\times1=2

eq-2\).

• When \(t=\frac{\pi}{2}\): \(2\sin(\frac{\pi}{2})+2\cos(\frac{\pi}{2})=2\times1+2\times0 = 2

eq-2\).

• When \(t=\pi\): \(2\sin(\pi)+2\cos(\pi)=2\times0+2\times(-1)=-2\).
• When \(t=\frac{3\pi}{2}\): \(2\sin(\frac{3\pi}{2})+2\cos(\frac{3\pi}{2})=2\times(-1)+2\times0=-2\).

Answer:

B. There is no solution.