Solve 2cos^2x + cosx − 1 = 0 for x over the interval [0, 2 π ).
1 answer:
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Answer:
option C is the correct answer
Step-by-step explanation:
System A: System B:
-x - 2y = 7 ........(1) -x - 2y = 7 .......(3)
5x - 6y = -3 ........(2) -16y= 32 ........(4)
Solution for system A: ( -3, -2)
For option A
Multiply the first equation by 3
3(- x - 2y) = 3(7) ⇒ - 3x - 6y = 21 ........... (5)
Add equation 2 and 5
5x - 6y = -3
<u> - 3x - 6y = 21</u>
2x -12y = 18 ⇒ 2( x - 6y ) = 2( 9 ) ⇒ x - 6y = 9
Option A is not equal to equation in system B so now we check option B.
For option B
Multiply the first equation by -5
-5(- x - 2y) = -5(7) ⇒ 5x + 10y = -35 ........... (6)
Add equation 2 and 6
5x - 6y = -3
<u> 5x + 10y = -35</u>
10x + 4y = -38 ⇒ 2( 5x + 2y ) = 2(-19) ⇒ 5x + 2y = -19
Option B is not equal to equation in system B so now we check option C.
For option C
Multiply the first equation by 5
5(- x - 2y) = 5(7) ⇒ - 5x - 10y = 35 ........... (7)
Add equation 2 and 7
5x - 6y = -3
<u> - 5x - 10y = 35</u>
0 - 16y = 32 ⇒ -16y = 32
Option C is equal to equation in system b so now we check if the solution to system B is same as system A.
Solution to system B:
First, we find value of y from equation 4:
-16y = 32 ⇒ y = ⇒ y = -2
Now, we put value of y in equation 3 to find value of x:
-x - 2y = 7 ⇒ -x - 2(-2) = 7 ⇒ -x + 4 = 7 ⇒ -x = 7 - 4 ⇒ -x = 3
multiply both sides by -1
-1 × (-x) = -1 × (3) ⇒ x = -3
Solution of system B = (-3, -2)
Solution of system B is the same as system A, so option C is correct.