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sin(x)+cos(x)<= 2

Solution

sin (x) + cos (x) \leq 2

True for all x \in R

Interval Notation

(- \infty, \infty)

Solution steps

sin (x) + cos (x) \leq 2

Use the following identity:

cos (x) + sin (x) = 2 sin (\frac{π}{4} + x)

2 sin (\frac{π}{4} + x) \leq 2

Divide both sides by

2

2 sin (\frac{π}{4} + x) \leq 2

Divide both sides by

2

\frac{2 sin ( \frac{π}{4} + x )}{2} \leq \frac{2}{2}

Simplify

\frac{2 sin ( \frac{π}{4} + x )}{2} \leq \frac{2}{2}

Simplify

\frac{2 sin ( \frac{π}{4} + x )}{2} : sin (\frac{π}{4} + x)

\frac{2 sin ( \frac{π}{4} + x )}{2}

Cancel the common factor:

2

= sin (\frac{π}{4} + x)

Simplify

\frac{2}{2} : 2

\frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{1}}{2 ^{\frac{1}{2}}} = 2^{1 - \frac{1}{2}}

= 2^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= 2

sin (\frac{π}{4} + x) \leq 2

sin (\frac{π}{4} + x) \leq 2

sin (\frac{π}{4} + x) \leq 2

Range of

sin (\frac{π}{4} + x) : - 1 \leq sin (\frac{π}{4} + x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (\frac{π}{4} + x) \leq 1

- 1 \leq sin (\frac{π}{4} + x) \leq 1

sin (\frac{π}{4} + x) \leq 2 and - 1 \leq sin (\frac{π}{4} + x) \leq 1 : - 1 \leq sin (\frac{π}{4} + x) \leq 1

Let

y = sin (\frac{π}{4} + x)

Combine the intervals

y \leq 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq 2 and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \leq 2

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

2 cos^{2} (x) + sin (2 x) \leq 0

\frac{1}{cos ( x )} \geq 2

cos (2 x) > \frac{1}{2}

sin (x) \geq cos (x)

cos (x) < \frac{3}{2}