All categories

3 years ago

Find all exact solutions on the interval [0, 2π). Look for opportunities to use trigonometric identities.

Mathematics

2 answers:

Nostrana [21]3 years ago

7 0

Answer:

x = π/2, 7π/6, or 11π/6

Step-by-step explanation:

sin² x − cos² x − sin x = 0

Use Pythagorean identity to write cosine in terms of sine.

sin² x − (1 − sin² x) − sin x = 0

sin² x − 1 + sin² x − sin x = 0

2 sin² x − sin x − 1 = 0

Factor:

(2 sin x + 1) (sin x − 1) = 0

Solve:

sin x = -1/2 or 1

x = π/2, 7π/6, or 11π/6

Ipatiy [6.2K]3 years ago

5 0

Answer:

pi/2, 7pi/6, 11pi/12

Step-by-step explanation:

sin²x - cos²x - sinx = 0

sin²x - (1 - sin²x) - sinx = 0

2sin²x - sinx - 1 = 0

2sin²x - 2sinx + sinx - 1 = 0

2sinx(sinx-1) + 1(sinx-1) =0

(2sinx + 1)(sinx - 1) = 0

sinx = -0.5, 1

sinx = 1 x = pi/2

sinx = -0.5 x = pi+pi/6 = 7pi/6

x = 2pi-pi/6 = 11pi/6

You might be interested in

Joe gave 1/4 of his total candies to his classmate then he gave 4/6 of when he had left to his brother when he went home then he

Makovka662 [10]

Answer:

112

Step-by-step explanation:

Given: Joe gave 1/4 of his total candies to his classmate.

Then, he gave 4/6 of when he had left to his brother.

He gave 25% of the remaining candies to his sister.

Finally, he only had 21 candies left.

Lets assume the total number of candies at the beginning be "x".

First, finding the number candies left after giving candies to classmate.

∴ Remaining candies= $x- x\times \frac{1}{4}$

Solving it to find remaining candies after giving candies to clasmate.

⇒ Remaining candies= $x-\frac{x}{4}$

Taking LCD as 4

⇒ Remaining candies= $\frac{4x-x}{4} = \frac{3x}{4}$

∴ Remaining candies after giving candies to clasmate= $\frac{3x}{4}$

now, finding the candies left after giving candies to his brother.

∴ Remaining candies= $\frac{3x}{4} - \frac{3x}{4} \times \frac{4}{6}$

Solving it to find the remaining candies after giving candies to his brother.

⇒ Remaining candies= $\frac{3x}{4} - \frac{x}{2}$

Taking LCD 4

⇒ Remaining candies= $\frac{3x-2x}{4} = \frac{x}{4}$

∴ Remaining candies after giving candies to his brother= $\frac{x}{4}$

We know, Joe was left with only 21 candies after giving candies to his sister.

Therefore, putting an equation for remaining candies to find the number of candies at the beginning.

⇒ $\frac{x}{4} - 25\% \times \frac{x}{4} = 21$

⇒ $\frac{x}{4} - \frac{0.25x}{4} = 21$

Taking LCD 4

⇒ $\frac{x-0.25x}{4} = 21$

⇒ $\frac{0.75x}{4} = 21$

Multiplying both side by 4

⇒ $0.75x= 21\times 4$

dividing both side by 0.75

⇒ $x= \frac{21\times 4}{0.75}$

∴ $x= 112$

Hence, Joe had 112 candies at the beginning.

6 0

3 years ago

Solve for m: (-11/6) + m = -2/9 M=

ss7ja [257]

Step-by-step explanation:

(-11/6) + m = -2/9

m = -2/9 + 11/6

the last common multiple of 9 and 6 is 18.

so, we are bringing both fractions to .../18.

m = -2×2/18 + 11×3/18 = -4/18 + 33/18 = 29/18

8 0

2 years ago

Please solve this ..................

AlladinOne [14]

Answer:

Step-by-step explanation:

√20=√5*2*2

√20=2√5

4 0

3 years ago

Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p

posledela

Answer:

P=0.0000037

P=0.00037%

Step-by-step explanation:

Probability

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

$\displaystyle \frac{4}{52}=\frac{1}{13}$

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

$\displaystyle \frac{3}{51}=\frac{1}{17}$

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

$\displaystyle \frac{2}{50}=\frac{1}{25}$

Finally, the probability of getting the remaining ace out of the 49 cards is:

$\displaystyle \frac{1}{49}$

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

$\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}$

$\displaystyle P= \frac{1}{270,725}$

P=0.0000037

P=0.00037%

3 0

2 years ago

There are 5 kids who want to start a game of Red Rover. Then 12 more kids come to play. If there are 2 teams in Red Rover, which

mixas84 [53]

Answer: The answer is one will have 8 and one will have 9 so roughly 8

Step-by-step explanation

So to start we need to add all the kids together that are playing

5+12= 17

Now we need to divide the number of kids by the number of teams that there are

17 divided by 2=8.5

So each team will have 8 but one will have nine

Hope this helps :D

7 0

2 years ago

Read 2 more answers