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2 years ago

10

Find the solutions to sin2(x) + cos(x) = 1, keeping 0 ≤ x < 2π

1 answer:

frozen [14]2 years ago

3 0

Sin2(x) +cos(x)=1
from the relation: (sin2(x) +cos2(x) =1 )
so , sin2(x)=1-cos2(x)
by subs. in the main eqn.
1-cos2(x) + cos(x) =1
by simplify the eqn.
cos(x) -cos2(x)=0
take cos(x) as a common factor
cos(x)* (1-cos(x))=0
then cos(x)=0 && cos(x)=1
cos(x)=0 if x= pi/2
& cos(x) = 1 if x = 0 , 2*pi
so the solution is x= {0,pi/2 , 2*pi}

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suppose you ask your three friends how many cups of coffee they each drink each week. they report the following: 3 cups, 6 cups,

Viktor [21]

Answer: The mean is 9 cups, the median is 6 cups, and the total is 27.

Step-by-step explanation:

To calculate the mean, add up all values and divide by the amount of all the values: 3 + 6 + 18 = 27, 27/3 = 9. The median is the middle number, which is 6 in the set 3, 6, 18 (least to greatest). Finally the total is already 27. Hope this helps, please consider Brainliest (click on the crown).

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1 year ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

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

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

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

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

8 0

2 years ago

The base of a rectangle is 15cm the perimeter is at most 94 cm. Write and solve an inequality to find the possible heights of th

tino4ka555 [31]

15 x 6 = 90 + 4 = 94 that's what I would do

4 0

3 years ago

Need help 8387.02 + 744.8

vivado [14]

The answer would be <span>9131.82</span>

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2 years ago

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mr_godi [17]

Answer:

$\frac{1}{6}$

Step-by-step explanation:

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5 0

2 years ago