Can you find all solutions of each of the following equations? a) 2sinxcosx=cosx b) 2sin^2x-3+1=0

1 Answer
Dean R.
May 1, 2018

# x = pi/2 + pi k quad #

# x=\pi/6 + 2\pi k # or

# x = {5 pi}/6 + 2 pi k #

for integer #k#.

That's #{ 30^circ, 90^circ, 150^circ, 270^circ } + 360^circ k#

Explanation:

That's two questions.

# 2 sin x cos x = cos x #

# 2 sin x cos x - cos x = 0 #

# cos x ( 2 sin x - 1) = 0 #

#cos x = 0# or #2 sin x -1 = 0#

#cos x = 0# or #sin x = 1/2#

#cos x = 0# happens at #\pm 90^circ,# which we write

# x = pi/2 + pi k quad # for integer #k#

The second is of course one of the two triangles that trig question writers use over and over. #sin x=1/2# happens at #30^circ# and its supplement #150^circ.# We write

# x=\pi/6 + 2\pi k # or

# x = {5 pi}/6 + 2 pi k #

# 2sin^2x-3+1=0 #

I'm gonna guess this one was supposed to read

# 2sin^2x-3 sin x +1=0 #

We factor,

# (2 sin x - 1)(sin x - 1) = 0#

# sin x = 1/2 # or #sin x = 1 #

We just saw this first one,

# x=\pi/6 + 2\pi k # or # x = {5 pi}/6 + 2 pi k #

The second is the max of the sine,

# x = pi/2 + 2pi k #

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