Urgent calculus help

Question

Solve the following inequality. Express the exact answer in interval notation, restricting your attention to −𝜋 ≤ x ≤ 𝜋.

sin2(x) < 1/2

Answer 1

sin^2(x) < 1/2, with -pi<x<pi

-sqr2/2 < sinx < sqr2/2

x is an element of 3 intervals

in degrees

(-45,45), (135,180), (-180,-135)

or in radians

(-pi/4, pi/4), (3pi/4, pi), (-pi, -3pi/4)

this isn't a calculus problem

it's a trig problem, and trig is part of the precalculus course

Answer 2

sin²(x)<1/2

+/-sin(x)<√2/2

1.sin(x)<√2/2

and -sin(x)<√2/2 or

2.sin(x)>-√2/2

in domain -π<x<π

for 1.,,,x₁<π/4 ,,,,note-x₁>-π/4,,,,,x₂=π-x₁>π-π/4=3π/4

x₂>3π/4

for2.,,,,,x₃>-π/4,,,,note-x₃<π/4,,,,x₄=π-x₃<π-(-π/4)=5π/4=5π/4-2π=-3π/4

x₄<-3π/4

have x such x<π/4 and x>-π/4

and x such that x>3π/4 and x<-3π/4

Answer 3

The problem tells us that sin² x < 1/2.

But let's do a simpler problem first.

Suppose they told us that y² < 4.

Let's take square roots on both sides to find out what y is.

-2 < y < +2

Got that? y can be on either side of 0 (i.e., y can be positive or negative), but in either case it must be pretty close to 0. If y is beyond these limits (e.g., if y = +3 or y = -3), then it will not be true that y² < 4.

Now back to the sin² x < 1/2 in this problem.

Let's take square roots on both sides to find out what sin x is.

-√(1/2) < sin x < +√(1/2)

Got that? sin x can be on either side of 0 (i.e., sin x can be positive or negative), but in either case it must be pretty close to 0. If sin x is beyond these limits, then it will not be true that sin² x < 1/2.

So for what angles x do we have -√(1/2) < sin x < +√(1/2)?

Let's start with a simpler problem: for what angles x do we have plain old sin x = +√(1/2)?

After all, equality is simpler than "less than".

Sin is just "opposite over hypotenuse", so let's see if we can write the expression +√(1/2) as a fraction in the form of "something over something else".

+√(1/2) = 1/√2

So we have an opposite side of length 1, and a hypotenuse of length √2. (I knew that √2 is the hypotenuse because the hypotenuse is always the longest side.) The Pythagorean Theorem tells is that the third side of the triangle must be 1 (because 1² + 1² = (√2)²).

Can you recognize these three sides 1, 1, √2 as the three sides of a 45-45-90 right triangle?

So an angle of 45 degrees (i.e. π/4 radians) would have a sine of 1/√2.

And, by the way, an angle of -45 degrees (i.e., -π/4 radians) would have a sine of -1/√2.

Draw a sine curve from x = -π to x = π. (The problem says that we are interested in only the x's that are in this interval.) Can you see that we have the desired -1/√2 < sin x < 1/√2 in three subintervals within the larger interval x = -π to x = π? These three subintervals, in order of how easy they are to see, are

(-π/4, π/4) (i.e, (-45 degrees, 45 degrees))

(3π/4, π) (i.e., (135 degrees, 180 degrees))

(-π, -3π/4) (i.e., -(180 degrees, -135 degrees))

So we want the union of these three subintervals, which we can write as

(-π/4, π/4) ∪ (3π/4, π) ∪ (-π, -3π/4)

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