Solve the equation on the interval 0
1) 2cos^(2)x - 3cosx + 1 = 0
a) 0, pi/3, 2pi/3
b) pi/3, pi/2, 5pi/3
c) 0, pi/6, 11pi/6
d) 0, pi/3, 5pi/3
2) sin^(2)x = 5(cosx + 1)
a) 0
b) 3pi/2
c) pi
d) no solution
3) sinx + (squareroot3) cosx = -1
a) 3pi/2, pi/6
b) 0, 2pi/3
c) 3pi/2, 5pi/6
d) pi/2, 7pi/6
PLEASE HELP!!! Show work if possible. Thanks!!! Appreciate it :)
1) 2cos^(2)x - 3cosx + 1 = 0
a) 0, pi/3, 2pi/3
b) pi/3, pi/2, 5pi/3
c) 0, pi/6, 11pi/6
d) 0, pi/3, 5pi/3
2) sin^(2)x = 5(cosx + 1)
a) 0
b) 3pi/2
c) pi
d) no solution
3) sinx + (squareroot3) cosx = -1
a) 3pi/2, pi/6
b) 0, 2pi/3
c) 3pi/2, 5pi/6
d) pi/2, 7pi/6
PLEASE HELP!!! Show work if possible. Thanks!!! Appreciate it :)
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The first two are quadratic equations, but instead of a simple variable you have a trig function of x. In problem 1, it's easier to see the solution if you let y = cos x. Then the equation becomes:
2y^2 - 3y + 1 = 0 .... which easily factors to:
(2y - 1)(y - 1) = 0 .... which in turn is solved by:
y = 1 or 1/2
But y = cos x, so your full answer is all the solutions to cos x = 1 or cos x = 1/2 on the interval from 0 to 2*pi. I trust you know how to do simple inverse cosines like this?
Problem 2 is more disguised, but essentially the same. Rewrite (sin x)^2 as 1 - (cos x)^2
1 - (cos x)^2 = 5(cos x + 1) ... and again let y = cos x
1 - y^2 = 5(y + 1)
y^2 + 5y + 4 = 0
(y + 1)(y + 4) = 0
y = -1 or -4
So cos x = -1 (one solution on [0,2*pi]) or cos x = -4 (no solutions anywhere...|cos x| <= 1 always!).
Problem 3 can be solved in a couple of ways. You can rewrite as sqrt(3)*cos x = -1 - sin x, square both sides and rewrite (cos x)^2 as 1 - (sin x)^2 on the left. That gives a quadratic in (sin x) this time that you can solve as above. It's really easier to divide both sides by 2 to get:
(1/2) sin x + (sqrt(3)/2) * cos x = -1/2
...but 1/2 = cos (pi/3) and sqrt(3)/2 = sin(pi/3) are well-known functions, giving:
cos(pi/3)*sin x + sin(pi/3)*cos x = -1/2
The left side now is in the form of the sine of a sum:
sin(x + pi/3) = -1/2
x + p/3 = arcsin(-1/2)
x = arcsin(-1/2) - pi/3
The general arcsine of -1/2 is (7pi/6 + n*2pi) or (11pi/6 + n*2pi) for integer n. Subtract pi/3 to get:
x = (5pi/6 + n*2pi) or (3pi/2 + n*2pi)
The only solutions for x in [0,2pi] are when n=0.
Edit: I forgot....the general method for simplifying a*cos x + b*sin x is to assume that:
a = r*sin θ
b = r*cos θ
...for some number r>0 and some angle θ. It's easy to compute r = sqrt(a^2 + b^2). θ takes a bit more work to see that the absolute value must be arccos(b/r), with the same sign as a. (I like that rule better than arctan(a/b) because there's never a zero-divide problem and the sign rule is easier.)
Then, finally,
a*cos x + b*sin x = r(sin θ)(cos x) + r(cos θ)(sin x) = r*sin(x + θ)
...where r and θ are determined as above.
2y^2 - 3y + 1 = 0 .... which easily factors to:
(2y - 1)(y - 1) = 0 .... which in turn is solved by:
y = 1 or 1/2
But y = cos x, so your full answer is all the solutions to cos x = 1 or cos x = 1/2 on the interval from 0 to 2*pi. I trust you know how to do simple inverse cosines like this?
Problem 2 is more disguised, but essentially the same. Rewrite (sin x)^2 as 1 - (cos x)^2
1 - (cos x)^2 = 5(cos x + 1) ... and again let y = cos x
1 - y^2 = 5(y + 1)
y^2 + 5y + 4 = 0
(y + 1)(y + 4) = 0
y = -1 or -4
So cos x = -1 (one solution on [0,2*pi]) or cos x = -4 (no solutions anywhere...|cos x| <= 1 always!).
Problem 3 can be solved in a couple of ways. You can rewrite as sqrt(3)*cos x = -1 - sin x, square both sides and rewrite (cos x)^2 as 1 - (sin x)^2 on the left. That gives a quadratic in (sin x) this time that you can solve as above. It's really easier to divide both sides by 2 to get:
(1/2) sin x + (sqrt(3)/2) * cos x = -1/2
...but 1/2 = cos (pi/3) and sqrt(3)/2 = sin(pi/3) are well-known functions, giving:
cos(pi/3)*sin x + sin(pi/3)*cos x = -1/2
The left side now is in the form of the sine of a sum:
sin(x + pi/3) = -1/2
x + p/3 = arcsin(-1/2)
x = arcsin(-1/2) - pi/3
The general arcsine of -1/2 is (7pi/6 + n*2pi) or (11pi/6 + n*2pi) for integer n. Subtract pi/3 to get:
x = (5pi/6 + n*2pi) or (3pi/2 + n*2pi)
The only solutions for x in [0,2pi] are when n=0.
Edit: I forgot....the general method for simplifying a*cos x + b*sin x is to assume that:
a = r*sin θ
b = r*cos θ
...for some number r>0 and some angle θ. It's easy to compute r = sqrt(a^2 + b^2). θ takes a bit more work to see that the absolute value must be arccos(b/r), with the same sign as a. (I like that rule better than arctan(a/b) because there's never a zero-divide problem and the sign rule is easier.)
Then, finally,
a*cos x + b*sin x = r(sin θ)(cos x) + r(cos θ)(sin x) = r*sin(x + θ)
...where r and θ are determined as above.