SOLUTION: Please help me solve this equation!
(a)Solve approximately the equations:
(i)2 sin x + cos x = 1.5
(ii)2 sin x + cos x = 0
(b)(i)3 cos x - 4 sin x + 1 = 0
(ii) 3 cos x = 4 s
Algebra.Com
Question 870286: Please help me solve this equation!
(a)Solve approximately the equations:
(i)2 sin x + cos x = 1.5
(ii)2 sin x + cos x = 0
(b)(i)3 cos x - 4 sin x + 1 = 0
(ii) 3 cos x = 4 sin x
PLEASE REPLY ASAP!
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
I am learning something important from this problem (and from the wiki I found once I figured out how to ask the question).
Two of your problems (aii and bii) were easy to answer without much algebra and/or trigonometry.
The other two were clearly related, but required more algebra work.
There had to be a common strategy to solve all four problems,
and using trigonometric identities had to be the key,
but I could not see how to use trigonometric identities to get where I wanted to get.
A quick internet search led me to the common strategy to solve the problems.
EACH PROBLEM AS A SEPARATE STRUGGLE:
(a)(ii)
<---><---><--->
From there we know that (in radians, or in degrees) is an answer.
Of course we know that there are infinite answers, (in radians) apart from each other,
because is a periodic function with a period of (with expressed in radians).
We could say that all of our approximate answers can be expressed as
(with and measured in radians).
The same strategy can be used to solve (b)(ii).
(I found as a general solution).
(a)(i)
<-->--><-->
Calling we can re-write the equation above as
<-->
We solve for using the quadratic formula:
The approximate solutions are
and .
With those values of we set out to find .
For :
There are two values of between 0 and (or between and that yield .
One is in quadrant I and the other in quadrant II.
With ,
to get we need to have ,
so we are looking foir a quadrant II solution.
(in radians, or in degress) is a solution.
Another possible value is (in radians, ) ),
but that is in quadrant I, with a positive , and
does not satisfy .
For :
The solution (in radians, or in degrees), in quadrant I, with looks promising,
as can be .
In fact, it can be verified to satisfy .
On the other hand, , in quadrant II, while it also yields ,
it yields and does not satisfy .
The solutions highlighted above, are the solutions between 0 and ,
and in general, all solutions can be expressed as
or
(with and measured in radians).
The same strategy can be used to solve (b)(i).
(The general solutions I found are
or ).
A COMMON STRATEGY:
The left side of the equation in (a)(ii) is
It is a "linear combination of sine and cosine functions".
That is a periodic function, like sine and cosine.
I can see that its period is .
It must be possible to express it as a single trigonometric function,
maybe ,with two constants and ,
where the cosine function is shifted right by and dilated vertically by a factor .
Now, how could I use trigonometric identities to transform
and
into a function like ?
It required a lot of thinking, and on the Sunday morning after such a Saturday night, I did not trust my brain that much.
I just googled "linear combinations of sine and cosine functions",
and helped myself to someone else's thinking.
Trigonometric identities tell us that
so
So if a linear combination of sine and cosine functions,
is equivalent to ,
then for all values of .
That means that
--->
Although that gives you two choices for C,
it is a formula-driven, apparently less cumbersome, common strategy to solve all four problems.
Applying those formulas:
has --->
The angle in quadrant I that has
measures approximately (in radians).
Using and we would conclude that
We re-write the equations that and solve:
(a)(i) -->-->-->
-->-->-->
(a)(ii)-->-->-->
-->-->
has --->
The angle in quadrant IV that has
measures approximately (in radians).
Using and we would conclude that
We re-write the equations that and solve:
(b)(i) -->-->-->-->
-->-->
The two solutions are and
, which can be written as .
(b)(ii) <-->-->-->
That means -->-->
RELATED QUESTIONS
URGENT HELP NEEDED!
Please help me solve this equation!
(a)Solve approximately the... (answered by richwmiller)
Solve the following equations for 0deg <= x <= 360deg:
(i) cos 2x cos x = sin 4x sin x... (answered by ikleyn)
Please help me with this problem. I need to solve the equation for solutions over the... (answered by Gogonati)
Please help me solve this equation: {{{ cos^2(x)+2sin(x)cos(x)-sin^2(x)=0 }}} with the... (answered by Edwin McCravy)
Find all solutions of the following equation:
2 cos^2(x) - 3 cos(x) = 2 sin(x)cos(x) - (answered by jim_thompson5910)
Please help me solve this equations:
1. {2 sin x-1=0}
2. graph {f(x)=sin(x-2pi)-1}... (answered by stanbon)
please help me solve this equation: solve exactly: sin(x) + 4csc(x) + 5=0
I have... (answered by stanbon)
kindly help me with these equations...thank you....
(3 cos x + 7)(-2 sin x - 1) = 0... (answered by richwmiller)
please help me solve:... (answered by jim_thompson5910)