Question

I am so confused. I have an exam tomorrow! Please help!!! I need a step by step explanation of these problems.

Solve each equation for exact solutions in the interval [0, 2π]:

a) 2sinx - √3 = 0



b) 2sinxcosx = √2 sinx

c) 2sin2x = 1 - cosx

Answer 1

\[2sinx-\sqrt{3} = 0 \rightarrow 2sinx=\sqrt{3} \rightarrow sinx=\frac{\sqrt{3}}{ 2 } \]
\[sinx=\sin60^{0} \rightarrow x =60^{0}\]

Answer 2

b. 2sinxcosx = √2 sinx \[ \frac{ sinx cosx }{ sinx } = \frac{ \sqrt{2} }{ 2 } \rightarrow cosx = \frac{ 1 }{ \sqrt{2} } \]
\[cosx = \cos 45^{0} \rightarrow x = 45^{0}\]