How do you solve #cosx=-1/sqrt2#?

2 Answers
Dean R.
May 4, 2018

# x = \pm 120^circ + 360^circ k #

Explanation:

#cos x = cos a# has solutions #x= \pm a + 360^circ k quad # for integer #k#.

#cos x = -1/sqrt{2} = cos( 120^circ) #

# x = \pm 120^circ + 360^circ k #

Jim G.
May 4, 2018

#"see explanation"#

Explanation:

#"since "cosx<0" then x is in second/third quadrants"#

#x=cos^-1(1/sqrt2)=pi/4larrcolor(red)"related acute angle"#

#rArrx=pi-pi/4=(3pi)/4larrcolor(red)"second quadrant"#

#"or "x=pi+pi/4=(5pi)/4larrcolor(red)"third quadrant"#

#"due to the periodicity of the cosine the solutions will"#
#"repeat every "2pi#

#color(blue)"solutions are"#

#x=(3pi)/4+2npito(n inZZ)#

#"or "x=(5pi)/4+2npito(n inZZ)#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
21642 views around the world
You can reuse this answer
Creative Commons License