How do you solve for x in #sin2x+cosx=0#?

3 Answers
Anish M.
Apr 5, 2018

So #x=pi/2,(7pi)/6#

Explanation:

We know that #sin2x=2sinxcosx#

So this becomes,

#2sinxcosx+cosx=0#

#(2sinx+1)(cosx)=0#

So either #2sinx+1=0# so #sinx=-1/2# in which case #x=(7pi)/6#
Or, #cosx=0# in which case #x=pi/2#

So #x=pi/2,(7pi)/6#

Ahmed A.
Apr 5, 2018

#x=210 or -30 or 330#

Explanation:

#Sin(2x)=2sin(x)cos(x)#

#2sin(x)cos(x)=-cos(x)#

#sin(x)=-1/2#

#sin^-1(x)=x#

sin of x is negative value Only at the third and the forth quarter.

#x=210 or -30 or 330#

Nghi N
Apr 5, 2018

# pi/2; (7pi)/6; (3pi)/2; (11pi)/6# for interval #(0, 2pi)#

Explanation:

sin 2x + cos x = 0
2sin x.cos x + cos x = 0
cos x(2sin x + 1) = 0
either factor should be zero.
a. cos x = 0
Unit circle gives 2 solutions -->
#x = pi/2 + 2kpi#, and
#x = (3pi)/2 + 2kpi#.
b. 2sin x + 1 = 0 --> #sin x= - 1/2#
Trig table and unit circle give 2 solutions:
#x = - pi/6 + 2kpi#, or #x = (11pi)/6 + 2kpi# (co-terminal)
#x = pi - (-pi/6) = (7pi)/6 + 2kpi#

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