Solve the equation: \$sin2x + 2 cosx sinx = \sqrt3\$.

The given equation is

\$sin2x + 2 cosx sinx = \sqrt3\$

\$sin2x = 2 sinx cosx\$.

Now plug the formula in the given equation

\$2sinx cosx + 2 cosx sinx = \sqrt3\$

Let take the common term 2cos⁡(x)sin⁡(x) then simplify it

\$2sinx cosx( 1 + 1) = \sqrt3\$
\$ 2(2cosx sinx) = \sqrt3\$

Simplify the equation

\$2 sin(2x) = \sqrt3\$
\$sin(2x) = \sqrt(3) /2\$

Now, we need to find the angles 2x where the sine is equal to the \$\sqrt(3)/2\$. This occurs when \$2x = (2(pi))/3 and (pi)/3\$.

Now, we solve for x:

\$2x = (2pi)/3\$ and \$2x = (pi)/3\$

\$x = (pi)/3\$ and \$x = (pi)/6\$

So, the solutions for the equation are \$ pi/3 "and" pi/6 \$.

Both cos\$π/3​\$ and sin\$π/6​\$ equal \$\sqrt(3)/2\$​​, and their product satisfies \$cos(x)sin(x) = \sqrt(3)/4\$​​. So, both values satisfy the equation

\$ pi/3, pi/6 \$

\$pi/2, 0\$ is not correct because these values do not satisfy the equation \$sin⁡(2x) + 2cos⁡(x)sin⁡(x) = \sqrt(3)\$

\$ pi/2, 0\$

The solution set from solving the equation is not fully represented by the intervals \$π/4\$ and \$π/2\$

\$ pi/4, pi/2 \$

This solution is incorrect because cos⁡(2π) and sin⁡(π) are both 0, so their product would be 0, not stem[\sqrt(3)/4]

\$ 2pi, pi \$

Option

A

Can you summarize what you’ve understood in the above steps?