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Popular >

3sin^2(θ)+7cos^2(θ)>0

Solution

3 sin^{2} (θ) + 7 cos^{2} (θ) > 0

Solution

True for all θ \in R

Interval Notation

(- \infty, \infty)

Solution steps

3 sin^{2} (θ) + 7 cos^{2} (θ) > 0

Use the following identity:

cos^{2} (x) + sin^{2} (x) = 1

Therefore

cos^{2} (x) = 1 - sin^{2} (x)

3 sin^{2} (θ) + 7 (1 - sin^{2} (θ)) > 0

Simplify

3 sin^{2} (θ) + 7 (1 - sin^{2} (θ)) : - 4 sin^{2} (θ) + 7

3 sin^{2} (θ) + 7 (1 - sin^{2} (θ))

Expand

7 (1 - sin^{2} (θ)) : 7 - 7 sin^{2} (θ)

7 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 7, b = 1, c = sin^{2} (θ)

= 7 \cdot 1 - 7 sin^{2} (θ)

Multiply the numbers:

7 \cdot 1 = 7

= 7 - 7 sin^{2} (θ)

= 3 sin^{2} (θ) + 7 - 7 sin^{2} (θ)

Simplify

3 sin^{2} (θ) + 7 - 7 sin^{2} (θ) : - 4 sin^{2} (θ) + 7

3 sin^{2} (θ) + 7 - 7 sin^{2} (θ)

Group like terms

= 3 sin^{2} (θ) - 7 sin^{2} (θ) + 7

Add similar elements:

3 sin^{2} (θ) - 7 sin^{2} (θ) = - 4 sin^{2} (θ)

= - 4 sin^{2} (θ) + 7

= - 4 sin^{2} (θ) + 7

- 4 sin^{2} (θ) + 7 > 0

Move

7

to the right side

- 4 sin^{2} (θ) + 7 > 0

Subtract

7

from both sides

- 4 sin^{2} (θ) + 7 - 7 > 0 - 7

Simplify

- 4 sin^{2} (θ) > - 7

- 4 sin^{2} (θ) > - 7

Multiply both sides by

- 1

- 4 sin^{2} (θ) > - 7

Multiply both sides by -1 (reverse the inequality)

(- 4 sin^{2} (θ)) (- 1) < (- 7) (- 1)

Simplify

4 sin^{2} (θ) < 7

4 sin^{2} (θ) < 7

Divide both sides by

4

4 sin^{2} (θ) < 7

Divide both sides by

4

\frac{4 sin ^{2} ( θ )}{4} < \frac{7}{4}

Simplify

sin^{2} (θ) < \frac{7}{4}

sin^{2} (θ) < \frac{7}{4}

For

u^{n} < a

, if

n

is even then

- \frac{7}{4} < sin (θ) < \frac{7}{4}

\frac{7}{4} = \frac{7}{2}

\frac{7}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7}{2}

- \frac{7}{2} < sin (θ) < \frac{7}{2}

a < u < b

then

a < u and u < b

- \frac{7}{2} < sin (θ) and sin (θ) < \frac{7}{2}

- \frac{7}{2} < sin (θ) :

True for all

θ \in R

- \frac{7}{2} < sin (θ)

Switch sides

sin (θ) > - \frac{7}{2}

Range of

sin (θ) : - 1 \leq sin (θ) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (θ) \leq 1

- 1 \leq sin (θ) \leq 1

sin (θ) > - \frac{7}{2} and - 1 \leq sin (θ) \leq 1 : - 1 \leq sin (θ) \leq 1

Let

y = sin (θ)

Combine the intervals

y > - \frac{7}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > - \frac{7}{2} and - 1 \leq y \leq 1

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

y > - \frac{7}{2}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all θ

True for all θ \in R

sin (θ) < \frac{7}{2} :

True for all

θ \in R

sin (θ) < \frac{7}{2}

Range of

sin (θ) : - 1 \leq sin (θ) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (θ) \leq 1

- 1 \leq sin (θ) \leq 1

sin (θ) < \frac{7}{2} and - 1 \leq sin (θ) \leq 1 : - 1 \leq sin (θ) \leq 1

Let

y = sin (θ)

Combine the intervals

y < \frac{7}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < \frac{7}{2} and - 1 \leq y \leq 1

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

y < \frac{7}{2}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all θ

True for all θ \in R

Combine the intervals

True for all θ \in R and True for all θ \in R

Merge Overlapping Intervals

True for all θ \in R and True for all θ \in R

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

True for all

θ \in R

and

True for all

θ \in R

True for all θ \in R

True for all θ

True for all θ \in R

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