Question 1066743
{{{sin(x)+cos(x)=0.4}}}
Let us square both sides of the equal sign.
The solutions of the resulting equation
will include all the solutions of the original equation,
and may include extra (extraneous) solutions,
but we can check at the end to eliminate the extraneous solutions.
{{{(sin(x)+cos(x))^2=0.4^2}}}
{{{sin^2(x)+cos^2(x)+2sin(x)cos(x)=0.16}}}
Using the trigonometric identity {{{sin^2(x)+cos^2(x)=1}}} ,
we can simplify the equation above to
{{{1+2sin(x)cos(x)=0.16}}}
{{{2sin(x)cos(x)=0.16-1}}}
Using the trigonometric identity {{{sin(2x)=2sin(x)cos(x)}}} ,
we can simplify the equation above to
{{{sin(2x)=-0.84}}}
Using the inverse function we find an approximate value for {{{2x}}} :
{{{2x=-57.14^o}}} (rounded).
There are many other angles that have {{{sin(2x)=-0.84}}} .
To begin with, for any angle {{{theta}}} , {{{sin(theta)=sin(180^o-theta)}}} .
Besides that, adding or subtracting any multiple of {{{360^o}}}
will give you a co-terminal angle with the same value for all its trigonometric functions.
So, {{{2x=180^o-(-57.14^o)=180^o+57.14^o=237.14^o}}} also has {{{sin(2x)=-0.84}}} ,
and so do all {{{2x}}} angles differing from {{{-57.14^o}}} or {{{237.14^o}}} by a multiple of {{{360^o}}} .
So, the solutions to the original equation would be among
{{{(-57.14^o)/2=-28.57^o}}} , {{{(360^o-57.14^o)/2=302.86^o/2=151.43^o}}} , {{{237.14^o/2=118.57^o}}} , {{{(360^o+237.14^o)/2=597.14^o/2=298.57^o}}} ,
and other angles differing by a multiple of {{{360^o}}} .
{{{highlight(x=-28.57^o)}}} and {{{highlight(x=118.57^o)}}} look like solutions,
and they check when substituted into the original equation.
So do all the angles differing from one of those solutions by multiples of {{{360^o}}} .
On the other hand,
{{{x=151.43^o}}} with {{{135^o<x<180^o}}} , and consequently
{{{-cos(x)>sin(x)>0}}} is obviously an extraneous solution,
since it will yield {{{sin(x)+cos(x)<0}}} .
So is {{{x=298.57^o}}} , with {{{270^o<x<315^o}}} and {{{-sin(x)>cos(x)>0}}} .