5sinx=5sqrt3cosx 0

Question

5sinx=5sqrt3cosx     0<x<2pi
What are all possible solutions?

Emery Adkins · Accepted Answer

solve #5sin x = 5sqrt3cos x#

Ans: #pi/3 and (4pi)/3# There are two ways. 1. First way. Divide both side by 5cos x (condition #cos x != 0, or x != pi/2 and x !=3pi/2#) #tan x = sqrt3# --> #x = pi/3# and #x = pi/3 + pi = (4pi)/3# 2. Second way. Simplify both sides by 5 #sin x - sqrt3cos x = 0# Replace in the equation (sqrt3) by #tan (pi)/3 = (sin (pi/3)/(cos pi/3))# #sin x.cos ((pi)/3) - sin ((pi)/3).cos x = 0# #sin (x - pi/3) = 0# #(x - pi/3) = 0 --># x = pi/3# and #(x - pi/3) = pi# --> #x = pi + pi/3 = (4pi)/3#

Madelyn Adams · Answer

undefined To find all possible solutions to the equation $5\sin(x) = 5\sqrt{3}\cos(x)$ for $0 < x < 2\pi$, we can rewrite the equation using trigonometric identities and then solve for $x$. First, divide both sides of the equation by 5 to simplify:

$$\sin(x) = \sqrt{3}\cos(x)$$

Now, we can use the identity $\cos(x) = \sin(\frac{\pi}{2} - x)$ to rewrite the equation:

$$\sin(x) = \sqrt{3}\sin\left(\frac{\pi}{2} - x\right)$$

Since $0 < x < 2\pi$, the values of $x$ and $\frac{\pi}{2} - x$ will lie in the first and fourth quadrants, respectively. In these quadrants, the sine function is positive. Therefore, we can drop the absolute value signs:

$$x = \frac{\pi}{2} - x$$

Now, solve for $x$:

$$x + x = \frac{\pi}{2}$$
$$2x = \frac{\pi}{2}$$
$$x = \frac{\pi}{4}$$

So, the possible solution for $x$ is $x = \frac{\pi}{4}$ within the given interval $0 < x < 2\pi$.

5sinx=5sqrt3cosx 0&lt;x&lt;2pi What are all possible solutions?

5sinx=5sqrt3cosx 0<x<2pi What are all possible solutions?