Question

Solve: sinxcosx=root3/4 Please? :)

Answer 1

hint: use the identity

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

Answer 2

\[\frac{ 1 }{ 2 }\sin(2x)=\sqrt[3]{4}\]

Answer 3

try to work it out from there

Answer 4

is it cube root 4 or root of 3/4

Answer 5

square root of 3 over 4

Answer 6

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

\(2\sin(x)\cos(x) = \dfrac{\sqrt{3}}{2}\)

\(\sin(2x) = \dfrac{\sqrt{3}}{2}\)

\(2x = \pi/3\;or\;2\pi/3\)

\(x = \pi/6\;or\;\pi/3\)

Answer 7

Thank you!!

Answer 8

Given sinxcosx=root3/4
divide the equation by cosx
\[\frac{sinx cosx}{cosx} = \frac{\sqrt{\frac{3}{4}}}{cosx} {sinx} = \frac{\sqrt{\frac{3}{4}}}{cosx} {sinx} = \frac{\sqrt{3}}{2 cosx} {sinx} - \frac{\sqrt{3}}{2 cosx} = 0\]
multiply with denominator
\[sinx.2cosx - \frac{ \sqrt{3} }{2 cosx } =0\]
\[sinx. 2 cosx -\sqrt{3} = 0\]
by making use of sin2x = sinx.2cosx
\[\sin2x - \sqrt{3} = 0\]
\[\sin2x = \sqrt{3}\]