Solve #cosx+sinx=2/3# ?

4 Answers
Monzur R. · Zor Shekhtman
Oct 4, 2016

Assuming you meant #sin#(#x#) #+# #cos#(#x#) #=# #2/3#,
there are two answers:

#cos(x_1) = 0.96#
#cos(x_2) = -0.29# (to 2dp.)

Explanation:

To solve this equation, we have to use trigonometric functions to isolate one of the unknowns and solve for that unknown.

In this case, we will use #sin^2#(#x#) #+# #cos^2#(#x#) = 1 and 2#sin(x)##cos(x)# = #sin(2x)#.

#sin#(#x#) #+# #cos#(#x#) = #2/3#

#(sin(x) + cos(x))^2# #=# #(2/3)^2 = # #4/9#

#sin^2(x)# #+# #cos^2(x)# #+# 2#sin(x)##cos(x)# #=# #4/9#

#1# #+# #sin(2x)# #=# #4/9#

#sin(2x)# #=# #- 5/9#

The function #sin(x)# is negative in the 3rd and 4th quadrants and #sin(theta)=sin(180^"o"-theta)#. Therefore, we have two solutions for #2x#:

#2x_1# #=# #sin^-1(-5/9)# #=# #-33.75^"o"#
#2x_2 = 180^"o"-(-33.75^"o") = 213.75^"o"#

From this:
#x_1# #=# #-33.75^"o"/2# #=# #-16.87^"o"#
#x_2 = 213.75^"o"/2 = 106.87^"o"#

#cos(x_1)=cos(-16.87^"o")# #=# #0.96#
#cos(x_2)=cos(106.87^"o")=-0.29#

Bdub
Oct 5, 2016

#x=1.87#-Round to 2 decimal places

Explanation:

#sinx+cosx=2/3#

Let's first use linear combination of cosine and sine with equal arguments formula to simplify it.

That is, we want to express
A cos x +B sin x in the form C cos (x-D). Note that A is the coefficient of cos x and B is the coefficient of sine x.

To find C use pythagorean formula and to find D we use one of these two formulas
#cos D = A/C, sinD=B/C#

From the given equation A = 1 and B = 1. So let's find C using pythagorean theorem.
#C=sqrt(A^2+B^2) = sqrt(1^2+1^2)=sqrt 2#

To find D we need to first figure out which quadrant x is in and because both cos x and sin x are positive it means that x is in quadrant one.

#cos D=1/sqrt2#
#D=cos^-1 (1/sqrt2) = pi/4#

Therefore #sinx+cosx=sqrt2 cos (x-pi/4)#

Now let's use that to solve the problem. That is,

#sqrt2 cos (x-pi/4)=2/3#

#cos (x-pi/4)=2/(3sqrt2)#

#x-pi/4=cos^-1(2/(3sqrt2))#

#x=cos^-1(2/(3sqrt2)) + pi/4#

#x=1.87#-Round to 2 decimal places

Cesareo R.
Oct 5, 2016

#x = 1/2 (2/3 pm sqrt[14]/3)#

Explanation:

Making #y = sin(x)# and #alpha = 2/3#

#y+sqrt(1-y^2) = alpha#
#1-y^2=(alpha-y)^2#
#1-y^2=alpha^2-2alpha y+y^2#
#2y^2-2alpha y+alpha^2-1=0# solving for #y#

#y= 1/2(alpha pm sqrt(2-alpha^2))#

#sin(x) = 1/2 (2/3 pm sqrt[14]/3)#

P dilip_k
Oct 5, 2016

Considering the given condition is

#cosx+sinx=2/3......(1)#

Now

#(cosx-sinx)^2+(cosx+sinx)^2=2(cos^2x+sin^2x)=2#

#=>(cosx-sinx)^2+(2/3)^2=2#

#=>(cosx-sinx)^2+(2/3)^2=2#

#=>(cosx-sinx)^2=2-4/9=14/9#

#=>(cosx-sinx)=+-sqrt14/3.....(2)#

Adding (1) and (2) we get

#2cosx=2/3+-sqrt14/3#

#=>cosx=1/2(2/3+-sqrt14/3)#

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