Answer:
Step-by-step explanation:
This question is asking us to find where sin(2x + 30) has a sin of 1. If you look at the unit circle, 90 degrees has a sin of 1. Mathematically, it will be solved like this (begin by taking the inverse sin of both sides):
![sin^{-1}[sin(2x+30)]=sin^{-1}(1)](https://tex.z-dn.net/?f=sin%5E%7B-1%7D%5Bsin%282x%2B30%29%5D%3Dsin%5E%7B-1%7D%281%29)
On the left, the inverse sin "undoes" or cancels the sin, leaving us with
2x + 30 = sin⁻¹(1)
The right side is asking us what angle has a sin of 1, which is 90. Sub that into the right side:
2x + 30 = 90 and
2x = 60 so
x = 30
You're welcome!
Option a: 
is the equivalent expression.
Explanation:
The expression is 
where 
Let us simplify the expression, to determine which expression is equivalent from the four options.
Multiplying the powers, we get,
Cancelling the like terms, we have,
This equation can also be written as,
Multiplying the terms in denominator, we have,
Thus, the expression which is equivalent to is
Hence, Option a is the correct answer.
Answer:
x>9
Step-by-step explanation:
Let A = {1, 2, 8} and B = {2, 7}.
P ( A ∪ B ) = {1, 2, 7, 8}
= 4/10
= 0.4
Answer: 0.4
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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. =)
