Question

Which of the following are solutions to the equation sinxcosx=sqrt(3)/4?

1. pi/12 +npi
2. pi/6 +npi
3. pi/12 +npi/2
4. 2pi/6 +npi

Answer 1

Multiply both sides by 4 to get

\(4\sin(x)\cos(x) = \sqrt{3}\)

Then square both sides:
\((4\sin(x)\cos(x))^2 = (\sqrt{3})^2\)

Answer 2

\(16\sin^2x\cos^2x = 3\)

Answer 3

Let \(\cos^2x = 1 - \sin^2x\):

\(16\sin^2x(1 - \sin^2x) = 3\)

Distribute:
\(16\sin^2x - 16\sin^4x = 3\)

Rearrange the equation in to the form \(ax^2 + bx + c = 0\)

\(-16\sin^4x + 16\sin^2x = 3\)

\(-16\sin^4x + 16\sin^2x - 3 = 0\)

Answer 4

Let \(\sin^2x = y\)

\(-16y^2 + 16y - 3 = 0\)

\(16y^2 - 16y + 3 = 0\)

Solve for y, then substitute \(\sin^2x\) back in to isolate x. Make sure to check your answers.

Answer 5

Thank you!

Answer 6

If you run into any problems, let me know.