Question

2sin(4x)+6=5

Answer 1

Let's solve the sine equation.

1. Express sine function in the left side of equation:

2\sin (4x)+6=5,\\ 2\sin (4x)=5-6,\\ 2\sin (4x)=-1,\\ \\ \sin (4x)=-\dfrac{1}{2}  .

2. Use the genereal solution to get the solution of your equation:

4x=(-1)^n\arcsin \left(-\dfrac{1}{2}\right) +2\pi n , where  n\in Z .

3. Find  \arcsin \left(-\dfrac{1}{2}\right) :

\arcsin \left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{3}  .

4. Substitute part 3 into part 2 and express x:

4x=(-1)^n \left(-\dfrac{\pi}{3}\right) +2\pi n , where  n\in Z,

x=(-1)^{n+1}\cdot \dfrac{\pi}{12}+\dfrac{\pi n}{2} , where  n\in Z.

5. Solutions of your equation are:

x=(-1)^{n+1}\cdot \dfrac{\pi}{12}+\dfrac{\pi n}{2} , where  n\in Z .

Answer 2

Let's solve the sine equation.

1. Express sine function in the left side of equation:

$2\sin (4x)+6=5,\\ 2\sin (4x)=5-6,\\ 2\sin (4x)=-1,\\ \\ \sin (4x)=-\dfrac{1}{2}$.

2. Use the genereal solution to get the solution of your equation:

$4x=(-1)^n\arcsin \left(-\dfrac{1}{2}\right) +2\pi n$, where $n\in Z$.

3. Find $\arcsin \left(-\dfrac{1}{2}\right)$:

$\arcsin \left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{3}$.

4. Substitute part 3 into part 2 and express x:

$4x=(-1)^n \left(-\dfrac{\pi}{3}\right) +2\pi n$, where $n\in Z,$

$x=(-1)^{n+1}\cdot \dfrac{\pi}{12}+\dfrac{\pi n}{2}$, where $n\in Z$.

5. Solutions of your equation are:

$x=(-1)^{n+1}\cdot \dfrac{\pi}{12}+\dfrac{\pi n}{2}$, where $n\in Z$.