Please also help with these... Thanks So much
a) 2sin²x = 1 + cosx
b) 2tan²x = secx - 1
c) (sinx)(cosx) = (cosx)
a)
2sin²x = 1 + cosx
=> 2( 1 - cos²x) = 1 + cosx
=> 2 - 2cos²x = 1 + cosx
=> 2cos²x + cosx - 1 = 0
=> (2 cos x - 1)(cos x + 1) = 0
cos x = -1 and 1/2
x = π/3, π and 5π/3 in the interval [0, 2π ]
b)
2tan²x = secx - 1
=> 2( sec²x - 1) = secx - 1
=> 2sec²x - 2 = secx - 1
=> 2sec²x - secx - 1 = 0
=> (2 sec x + 1) (sec x - 1) = 0
sec x = -1/2 and 1
only valid solution is sec x = 1
x = 0 and 2π
c)
(sinx)(cosx) = (cosx)
=> cos x ( sin x - 1) = 0
cos x = 0 and sin x = 1
x = π/2, 3π/2 and π/2
a) 2 sin² (x) = 1 + cos (x)
=> 2 ( 1 - cos² (x) ) = 1 + cos (x)
=> 2 - 2 cos² (x) = 1 + cos (x)
=> 1 - 2 cos² (x) - cos (x) = 0
=> -2 cos² (x) - cos (x) + 1 = 0
Let y = cos (x)
Then,
-2 cos²(x) - cos (x) + 1 = 0
becomes
-2y² - y + 1 = 0
-2y² - 2y + y + 1 = 0
(-2y² - 2y) (y+ 1) = 0
-2y (y + 1) (y + 1) = 0
( -2y+ 1) (y + 1) = 0
Now, substituting in cos (x) for y,
( -2y + 1) (y + 1) = 0
( -2 cos (x) + 1) ( cos (x) + 1) = 0
set each factor equal to 0 and solve for x:
-2 cos (x) + 1 = 0 <-- let's call this equation (1)
cos (x) + 1 = 0 <-- let's call this equation (2)
From equation (1),
cos (x) = ½
From the unit circle,
x = {60 + 360n , 300 + 360n} where n is any integer. I need to specify that because you didn't give me the restrictions for x.
From equation (2),
cos (x) = -1
From the unit circle again,
x = {180 + 360n} where n, once again, is any integer
You can solve b) and c) in similar manner.
Hope this helps!!
a) Use sin^2 + cos^2 =1 -> sin^2 = 1-cos^2
2(1-cos^2) = 1 + cos
2 - 2 cos^2 = 1 + cos
2 cos^2 + cos - 1 = 0
(2cos - 1)( cos + 1 = 0)
cos x = 1/2 or cos x = -1
So x = 30 or 180
b) tan = sin/cos sec = 1/cos
2 (sin/cos)^2 = 1/cos - 1
Multiply by cos^2
2 sin^2 = cos - cos^2
Using sin^2 = 1-cos^2
2 - 2cos^2 = cos - cos^2
Cos^2 + cos - 2 = 0
(cos + 2)(cos - 1) = 0
cos x = 1 so x = 0 or 180
c) Cancel cos leaving sin x = 1 x = 90
Cos x = 0 -> x = 90 or 270
Learn or learn to derive your identities.
http://en.wikipedia.org/wiki/List_of_trigonometric...
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Answers & Comments
a)
2sin²x = 1 + cosx
=> 2( 1 - cos²x) = 1 + cosx
=> 2 - 2cos²x = 1 + cosx
=> 2cos²x + cosx - 1 = 0
=> (2 cos x - 1)(cos x + 1) = 0
cos x = -1 and 1/2
x = π/3, π and 5π/3 in the interval [0, 2π ]
b)
2tan²x = secx - 1
=> 2( sec²x - 1) = secx - 1
=> 2sec²x - 2 = secx - 1
=> 2sec²x - secx - 1 = 0
=> (2 sec x + 1) (sec x - 1) = 0
sec x = -1/2 and 1
only valid solution is sec x = 1
x = 0 and 2π
c)
(sinx)(cosx) = (cosx)
=> cos x ( sin x - 1) = 0
cos x = 0 and sin x = 1
x = π/2, 3π/2 and π/2
a) 2 sin² (x) = 1 + cos (x)
=> 2 ( 1 - cos² (x) ) = 1 + cos (x)
=> 2 - 2 cos² (x) = 1 + cos (x)
=> 1 - 2 cos² (x) - cos (x) = 0
=> -2 cos² (x) - cos (x) + 1 = 0
Let y = cos (x)
Then,
-2 cos²(x) - cos (x) + 1 = 0
becomes
-2y² - y + 1 = 0
-2y² - 2y + y + 1 = 0
(-2y² - 2y) (y+ 1) = 0
-2y (y + 1) (y + 1) = 0
( -2y+ 1) (y + 1) = 0
Now, substituting in cos (x) for y,
( -2y + 1) (y + 1) = 0
becomes
( -2 cos (x) + 1) ( cos (x) + 1) = 0
set each factor equal to 0 and solve for x:
-2 cos (x) + 1 = 0 <-- let's call this equation (1)
cos (x) + 1 = 0 <-- let's call this equation (2)
From equation (1),
cos (x) = ½
From the unit circle,
x = {60 + 360n , 300 + 360n} where n is any integer. I need to specify that because you didn't give me the restrictions for x.
From equation (2),
cos (x) = -1
From the unit circle again,
x = {180 + 360n} where n, once again, is any integer
You can solve b) and c) in similar manner.
Hope this helps!!
a) Use sin^2 + cos^2 =1 -> sin^2 = 1-cos^2
2(1-cos^2) = 1 + cos
2 - 2 cos^2 = 1 + cos
2 cos^2 + cos - 1 = 0
(2cos - 1)( cos + 1 = 0)
cos x = 1/2 or cos x = -1
So x = 30 or 180
b) tan = sin/cos sec = 1/cos
2 (sin/cos)^2 = 1/cos - 1
Multiply by cos^2
2 sin^2 = cos - cos^2
Using sin^2 = 1-cos^2
2 - 2cos^2 = cos - cos^2
Cos^2 + cos - 2 = 0
(cos + 2)(cos - 1) = 0
cos x = 1 so x = 0 or 180
c) Cancel cos leaving sin x = 1 x = 90
Cos x = 0 -> x = 90 or 270
Learn or learn to derive your identities.
http://en.wikipedia.org/wiki/List_of_trigonometric...
http://en.wikipedia.org/wiki/List_of_trigonometric...