SOLUTION: How do you solve sin x + square root of 2 = - sin x on the interval -pi/2 < x < pi/2

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Question 1206316: How do you solve sin x + square root of 2 = - sin x on the interval -pi/2 < x < pi/2
Found 4 solutions by MathLover1, mananth, ikleyn, math_tutor2020:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

on the interval






Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!

sin x + square root of 2 = - sin x
sin x + square root of 2 + sin x=0
2sin x =- sqrt( 2 )
sin x = -sqrt(2)/2
sin (x)= - 1/sqrt(2)
the interval -pi/2 < x < pi/2
sine is negative in the third and fourth quadrants, find the angle in the third quadrant.
x =-pi/2





Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.

Be careful :   the answer   x = -pi/2  by  @mananth is  INCORRECT.

The correct answer is   x = -pi/4.

This angle is in the fourth quadrant (not in the third quadrant, as @mananth mistakenly states).



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

We can use a spreadsheet to check each answer mentioned (x = -pi/2 and x = -pi/4)
xLHSRHSLHS = RHS?
-pi/20.4142141No
-pi/40.7071070.707107Yes

The table shows that only x = -pi/4 works.
LHS = left hand side
RHS = right hand side

Furthermore,
-pi/2 < -pi/4 < pi/2
Multiply all sides by -4
2pi > pi > -2pi
which flips to
-2pi < pi < 2pi

The last inequality
-2pi < pi < 2pi
being true leads back to
-pi/2 < -pi/4 < pi/2
also being true

Or you could use these decimal approximations
pi/2 = 1.570796
pi/4 = 0.785398
So
-pi/2 < -pi/4 < pi/2
becomes
-1.570796 < -0.785398 < 1.570796

--------------------------------------------------------------------------

Another way to verify the answer is to graph h(x) = f(x) - g(x)
where
f(x) = sin(x) + sqrt(2)
g(x) = -sin(x)

Or basically you need to graph h(x) = 2*sin(x)+sqrt(2)
https://www.desmos.com/calculator/s6agjxpf7d
Desmos is free graphing software. GeoGebra is also another good choice.
The curve intersects the x axis at (-pi/4, 0) to confirm that x = -pi/4 is the only solution to this equation on this specified domain interval.

If the -pi/2 < x < pi/2 portion wasn't required, then there would be infinitely many solutions.
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