Find the coefficient $c$ so the equation is even

Question

Hello so I have been solving a task and I dont know how to solve it completely.The task says: Find the $c$ so that function $f(x) = 3\sin(2x +c) + 3$ is even. I know that if the function is odd this must be valid $3\sin(-2x + c) + 3 = 3\sin(+2x + c) + 3$ But i dont know how to find $c$?

Thank you!

Answer 1

You made up the equation correctly. Now you must solve it to find $c$: $$3\sin(-2x + c) + 3 = 3\sin(+2x + c) + 3 \Rightarrow \\ \sin (-2x)\cos c+\sin c\cos (-2x)=\sin (2x)\cos c+\sin c\cos (2x) \Rightarrow \\ 2\sin 2x\cos c=0 \Rightarrow \\ \cos c=0 \Rightarrow \\ \boxed{c=\frac{\pi}{2}+k\pi, k\in \mathbb Z}$$ Note: It was used the addition rule: $$\sin(a+b)=\sin a\cos b+\cos a\sin b.$$ You can also use sum to product rule: $$\sin a-\sin b=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}.$$

Answer 2

Hint:

$$\sin(t)=\cos(\frac\pi2 - t)$$ And $\cos(x)$ is an even function.

Answer 3

Even function: $f(-x)=f(x)$, i.e

$\sin (-2x+C)=\sin (2x+C)$;

1) $-2x+C = 2x+ C +2πk, k \in \mathbb{Z}.$

No condition for $C$.

$\small{2) -2x+C= (π - (2x +C)) +2πk, k \in \mathbb{Z}.}$

$2C= π +2πk$ , $k \in \mathbb{Z}.$

$C=π/2 +πk$, $k \in \mathbb{Z}$

Recall:

$\sin a = \sin b$ implies

1) $a= b +2πk$, $k \in \mathbb{Z}$, or

2) $a = (π-b) +2πk$, $k \in \mathbb{Z}$.