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Solve the trigonometric equation:

Restriction for the solution:

Square both sides of
(i):

![\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2%5Ccdot%20%281-sin%5E2%5C%2Cx%29-sin%5C%2Cx%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2-2%5C%2Csin%5E2%5C%2Cx-sin%5C%2Cx%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B-%5C%2C%5Cdfrac%7Bsin%5C%2Cx%7D%7Bcos%5E2%5C%2Cx%7D%5Ccdot%20%5Cleft%5B2%5C%2Csin%5E2%5C%2Cx%2Bsin%5C%2Cx-2%20%5Cright%5D%3D0%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2Cx%5Ccdot%20%5Cleft%5B2%5C%2Csin%5E2%5C%2Cx%2Bsin%5C%2Cx-2%20%5Cright%5D%3D0%7D)
Let

So the equation becomes

Solving the quadratic equation:



You can discard the negative value for
t. So the solution for
(ii) is

Substitute back for
t = sin x. Remember the restriction for
x:

where
k is an integer.
I hope this helps. =)
The equation in the form y = mx + b is y = 0.35x + 36.
<h3>How to represent a linear equation?</h3>
The equation formed from the expression is a linear equation and should be expressed in the slope intercept form.
y = mx + b
where
Therefore,
y = mx + b
where
- x = the number of monthly minutes used
- y = the total monthly of the Next fell plan
using the coordinates, let's find m
(470, 200.5)(590, 242.5)
Hence,
m = 242.5 - 200.5 / 590 - 470
m = 42 / 120
m = 0.35
Therefore,
y = 0.35x + b
200.5 = 0.35(470) + b
200.5 - 164.5 = b
b = 36
Finally, the equation in the form y = mx + b is y = 0.35x + 36.
learn more on linear equation here: brainly.com/question/24223023
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We just use the sinus rule to calculate this
a / sin alpha = b / sin beta
Alpha is the 90 degree angle from the wall to the floor
beta is the angle from the top of the ladder to the wall,
wich is 90 degrees - 75 degrees (triangle has 180 degrees angles, one is 90 degrees), so beta = 15 degrees
14 foot / sin 90 = b / sin 15
sin 90 = 1
we move the sin15 over to the other side
14 foot x sin 15 = b
b = 3.62 foot
Answer:
1/10
Step-by-step explanation:
Decimal Form:
0.09
0.1
Fraction Form:
9/100
1/10
Percentage Form:
9%
10%
Hope that helps :-)
Answer:
12
Step-by-step explanation:
Since a die has six faces, 6 outcomes are possible
Since a coin has two faces, 2 outcomes are possible
Multiply the chances together and you get 12