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

3 Answers
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

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

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