Question

a. 0b. pi/6c. 5pi/6d. piThese are 4 options but there can be more than 2 or 3 correct answers.Find the solution of each equation the interval( 0, 2pi).

Answer 1

The answers are

a. 0

b. π/6

c. 5π/6

d. π

Step-by-step explanation:

If we replace each of these options into the equation we can see that all makes the equation true:

`\begin{gathered} \sin ^20-\sin 0+1=\cos ^20 \\ 0-0+1=1 \\ 1=1\text{ true} \end{gathered}`
`\begin{gathered} \sin ^2(\pi)/(6)-\sin (\pi)/(6)+1=\cos ^2(\pi)/(6) \\ (1)/(2^2)-(1)/(2)+1=(\frac{\sqrt[]{3}}{2})^2 \\ (1)/(4)-(1)/(2)+1=(3)/(4) \\ 1-(1)/(4)=(3)/(4) \\ (3)/(4)=(3)/(4)\text{ true} \end{gathered}`
`\begin{gathered} \sin ^2(5\pi)/(6)-\sin (5\pi)/(6)+1=\cos ^2(5\pi)/(6) \\ (1)/(2^2)-(1)/(2)+1=(-\frac{\sqrt[]{3}}{2})^2^{} \\ (1)/(4)-(1)/(2)+1=(3)/(4) \\ (3)/(4)=(3)/(4)\text{ true} \end{gathered}`
`\begin{gathered} \sin ^2\pi-\sin \pi+1=\cos ^2\pi \\ 0-0+1=(-1)^2 \\ 1=1\text{ true} \end{gathered}`