Question

What are all the solutions to the equation sin 2x = 2cos x in the interval [0, 2π)?

Answer 1

Answer:

$x=\dfrac{1}{2} \pi, \dfrac{3}{2} \pi \quad \textsf{in the interval}\:[0,2 \pi)$

Step-by-step explanation:

Rewrite the given equation so that it equals zero, using the trig identity:

$\sin(2x)=2\sin(x)\cos(x)$

$\begin{aligned}\sin(2x) & =2\cos(x)\\\sin(2x)-2\cos(x) & =0\\2\sin(x)\cos(x)-2\cos(x) & =0\\2\cos(x)[sin(x)-1] & =0\end{aligned}$

$\begin{aligned}\textsf{Therefore}:\quad 2\cos(x) & =0\\\cos(x) & =0\\x & =\dfrac{1}{2} \pi \pm 2 \pi n, \dfrac{3}{2} \pi \pm 2 \pi n\end{aligned}$

$\begin{aligned}\textsf{Therefore}:\quad \sin(x)-1 & =0\\\sin(x) & =1\\x & =\dfrac{1}{2} \pi \pm 2 \pi n\end{aligned}$

$\implies x=\dfrac{1}{2} \pi, \dfrac{3}{2} \pi \quad \textsf{in the interval}\:[0,2 \pi)$

Answer 2

An equation is formed of two equal expressions. The value of x for which the sin(2x) = 2cos(x) will have solution in the interval [0, 2π) is (π/2) and (3π/2).

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The equation can be rewritten as,

sin(2x) = 2 cos(x)

2 sin(x) cos(x) = 2 cos(x)

2 sin(x) cos(x) - 2 cos(x) = 0

Taking 2cos(x) as common,

2 cos(x) [sin(x) - 1] = 0

Now, the two factors can be written as,

cos(x) = 0

x = (π/2)±2πn, (3π/2)±2πn

sin(x) - 1 = 0

sin(x) = 1

x = (π/2)±2πn

Hence, the value of x for which the sin(2x) = 2cos(x) will have solution in the interval [0, 2π) is (π/2) and (3π/2).

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