SOLUTION: <b>Solve: sin<sup>2</sup><i>x</i>=sin<i>x</i>·cos<i>x</i>, 0 <font face = "symbol">£</font> x < 2<font face = "symbol">p</font> a. <i>x</i>=0, <font face = "symbol">p</font>/4

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Question 23294: Solve: sin2x=sinx·cosx, 0 £ x < 2p

a. x=0, p/4
b. x=p/4, 5p/4
c. x=0, 3p/4, p, 7p/4
d. x=0, p/4, p, 5p/4
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
Solve: sin2x = sinxcosx, 0 £ x < 2p
 
a. x=0, p/4
b. x=p/4, 5p/4
c. x=0, 3p/4, p, 7p/4
d. x=0, p/4, p, 5p/4

sin2x = sinx·cosx

sin2x - sinx·cosx = 0

sinx(sinx - cosx) = 0

Set each factor = 0

sinx = 0

x = 0, p  that's two solutions

sinx - cosx = 0

sinx = cosx

 sinx
—————— = 1
 cosx

tanx = 1

x = p/4, 5p/4

Solutions: 0, p/4, p, 5p/4, choice d.

Edwin
AnlytcPhil@aol.com
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