Solve the equation for x if 0 ≤ x < 2pi: 2 cos(x) + 3 tan(x) = 3 sec(x)

[gjbadt]

Solve the equation for x if 0 ≤ x < 2pi

2 cos(x) + 3 tan(x) = 3 sec(x)

My Work:

Rewrite tan(x) as sin(x)/cos(x) and rewrite sec(x) as 1/cos(x):
2cos(x) + 3sin(x)/cos(x) = 3/cos(x)

Then, multiply 2cos(x) by cos(x)/cos(x) to get a common denominator:
2cos^2(x)/cos(x) + 3sin(x)/cos(x) = 3/cos(x)

Multiply both sides by cos(x):
2cos^2(x) + 3sin(x) = 3

Using the trig identity cos^2(x) + sin^2(x) = 1, I substituted 1 - sin^2(x) for cos^2(x):
2(1 - sin^2(x)) + 3sin(x) = 3

Multiply that out:
2 - 2sin^2(x) + 3sin(x) = 3

Subtract 3 from both sides:
-2sin^2(x) + 3sin(x) - 1 = 0

Factor out a -1 from the left side:
-1(2sin^2(x) - 3sin(x) + 1) = 0

Factor the quantity 2sin^2(x) - 3sin(x) + 1 as (sin(x) - 1)(2sin(x) -1)
-1(sin(x) - 1)(2sin(x) -1) = 0

Distribute -1 (if I would have multiplied both sides by -1 as opposed to factoring out a -1, sin(x) would still equal 1 in the end):
(-sin(x) + 1)(2sin(x) -1) = 0

Set each term equal to 0
-sin(x) + 1 = 0 AND 2sin(x) -1 = 0

So,
sin(x) = 1 AND sin(x) = 1/2

Therefore,
x = 90 degrees (or pi/2) AND 30 degrees (or pi/6) AND 150 degrees (or 5pi/6)

However,
Because I multiplied both sides of the equation by cos(x), I assumed cos(x) did not equal 0. As a result, pi/2 must be thrown out because the cosine of pi/2 = 0

My solutions are pi/6 and 5pi/6

Apparently the solutions are 7pi/6 and 11pi/6, which leads me to believe I have a sign error along the way. Can someone help me out with this? Thanks.

 

[Deleted member 4993]

Solve the equation for x if 0 ≤ x < 2pi

2 cos(x) + 3 tan(x) = 3 sec(x)

My Work:

Rewrite tan(x) as sin(x)/cos(x) and rewrite sec(x) as 1/cos(x):
2cos(x) + 3sin(x)/cos(x) = 3/cos(x)

Then, multiply 2cos(x) by cos(x)/cos(x) to get a common denominator:
2cos^2(x)/cos(x) + 3sin(x)/cos(x) = 3/cos(x)

Multiply both sides by cos(x):
2cos^2(x) + 3sin(x) = 3

Using the trig identity cos^2(x) + sin^2(x) = 1, I substituted 1 - sin^2(x) for cos^2(x):
2(1 - sin^2(x)) + 3sin(x) = 3

Multiply that out:
2 - 2sin^2(x) + 3sin(x) = 3

Subtract 3 from both sides:
-2sin^2(x) + 3sin(x) - 1 = 0

Factor out a -1 from the left side:
-1(2sin^2(x) - 3sin(x) + 1) = 0

Factor the quantity 2sin^2(x) - 3sin(x) + 1 as (sin(x) - 1)(2sin(x) -1)
-1(sin(x) - 1)(2sin(x) -1) = 0

Distribute -1 (if I would have multiplied both sides by -1 as opposed to factoring out a -1, sin(x) would still equal 1 in the end):
(-sin(x) + 1)(2sin(x) -1) = 0

Set each term equal to 0
-sin(x) + 1 = 0 AND 2sin(x) -1 = 0

So,
sin(x) = 1 AND sin(x) = 1/2

Therefore,
x = 90 degrees (or pi/2) AND 30 degrees (or pi/6) AND 150 degrees (or 5pi/6)

However,
Because I multiplied both sides of the equation by cos(x), I assumed cos(x) did not equal 0. As a result, pi/2 must be thrown out because the cosine of pi/2 = 0

My solutions are pi/6 and 5pi/6

Apparently the solutions are 7pi/6 and 11pi/6, which leads me to believe I have a sign error along the way. Can someone help me out with this? Thanks.

for the given problem, and domain,

cos(x) + 3 tan(x) = 3 sec(x) domain 0≤ x ≤2π

Your answers are correct (satisfies the equations).

 

[gjbadt]

That's what I thought. Yet, when I e-mailed my professor his response was: "The answers you have are not correct. The right answers are 7pi/6 and 11pi/6". I'm beginning to think he doesn't know what he's talking about lol...

Thanks for the help!

 

[Deleted member 4993]

That's what I thought. Yet, when I e-mailed my professor his response was: "The answers you have are not correct. The right answers are 7pi/6 and 11pi/6". I'm beginning to think he doesn't know what he's talking about lol...

Thanks for the help!

Not so fast...

Make sure you are solving the correct problem.

For if the problem was:

Solve for x when: 2cos(x) - 3tan(x) = 3 sec(x) and 0 ≤ x < 2π

the answers would be 7π/6 and 11π/6