Trig identity and equations

  • #1
how can i do this trig identity

ln|sec(theta) + tan(theta| + ln|sec(theta) + tan(theta)| = 0

and this trig equation
4sin(squared)x + 2cos(squared)x = 3

and cos2(theta) + 3 = 5cos(theta)


any help will be appreciated
 

Answers and Replies

  • #2
dextercioby
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Rewrite the first.It doesn't make any sense.

For the second,use [itex] \cos^{2}x=1-\sin^{2}x [/itex] for transform it into a quadratic in "sin x".

For the third,use [itex] \cos 2x=2\cos^{2}x-1 [/tex] to transform it into a quadratic in "cos x".

Daniel.
 
  • #3
ln|sec(theta) + tan(theta)| + ln|sec(theta) - tan(theta)| = 0

| (line) is absolute value
 
  • #4
dextercioby
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There's no big deal

[tex] \ln\left|\sec\theta+\tan\theta\right|+\ln\left|\sec\theta-\tan\theta\right|=\ln|\sec^{2}\theta-\tan^{2}\theta|=\ln 1 =0 [/tex]

Q.e.d.

Daniel.
 
  • #5
uart
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needhelp123 said:
how can i do this trig identity

ln|sec(theta) + tan(theta| + ln|sec(theta) + tan(theta)| = 0

and this trig equation
4sin(squared)x + 2cos(squared)x = 3

and cos2(theta) + 3 = 5cos(theta)


any help will be appreciated

Every one of those equations is nonsense. Sure you need help, you need help to transcribe your questions accurately. You are doomed to failure in any problem if you cant even write the problem description properly.

BTW. Congrats to dextercioby for figuring out what you actually meant to write in at least one of those problems.
 
  • #6
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Practical Practice...


Trig identity confirmed:
[tex]\boxed{\ln\left|\sec\theta+\tan\theta\right|+\ln\left|\sec\theta-\tan\theta\right|=\ln|\sec^{2}\theta-\tan^{2}\theta|=\ln 1 =0} [/tex]

[tex] 4 \sin^2 x + 2 \cos^2 x = 3 [/tex]
[tex] \cos^2 x = 1 - \sin^2 x [/tex] - identity
[tex] 4 \sin^2 x + 2(1 - \sin^2 x) - 3 = 0 [/tex]
[tex] 4 \sin^2 x + 2 - 2 \sin^2 x - 3 = 0 [/tex]
[tex] 4 \sin^2 x - 2 \sin^2 x - 1 = 0 [/tex]
[tex] 2 \sin^2 x - 1 = 0 [/tex]
[tex] \sin^2 x = \frac{1}{2} [/tex]
[tex] \boxed{\sin x = \pm \frac{\sqrt{2}}{2}} [/tex]
:biggrin:

[tex] \cos 2x + 3 = 5 \cos x [/tex]
[tex] \cos 2x - 5 \cos x + 3 = 0 [/tex]
[tex] \cos 2x = 2 \cos^2 x - 1 [/itex] - identity
[tex] (2 \cos^2 x - 1) - 5 \cos x + 3 = 0 [/tex]
[tex] 2 \cos^2 x - 5 \cos x + 2 = 0 [/tex]
[tex] a = 2 \; \; b = -5 \; \; c = 2 [/tex]
[tex] \cos x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} [/tex]
[tex] \cos x = \frac{5 \pm \sqrt{9}}{4} [/tex]
[tex] \cos x = \frac{5 \pm 3}{4} [/tex]
[tex] \boxed{\cos x = \frac{1}{2}} [/tex]
[tex] \cos x = 2 \; \; \text{(no solution)} [/tex]
:biggrin:
 
Last edited:
  • #7
dextercioby
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U have to disregard [itex] \cos x=2 [/itex],okay?

Daniel.
 
  • #8
Zurtex
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dextercioby said:
U have to disregard [itex] \cos x=2 [/itex],okay?

Daniel.
Unless the question asks for complex solutions.
 
  • #9
dextercioby
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"x" typically depicts real variable...:wink:

If it were "z",it may have considered imaginary solutions,too...

Daniel.
 
  • #10
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Practice Problems...



It is my opinion that if the original equation states both [itex] \sin x [/itex] and [itex] \cos x [/itex] then both solutions must be located in order to lock on the solution quadrants, if possible.

[tex] 4 \sin^2 x + 2 \cos^2 x = 3 [/tex]
[tex] \sin^2 x = 1 - \cos^2 x [/tex] - identity
[tex] 4 (1 - \cos^2 x) + 2 \cos^2 x - 3 = 0 [/tex]
[tex] 4 - 4 \cos^2 x + 2 \cos^2 x - 3 = 0 [/tex]
[tex] -2 \cos^2 x + 1 = 0 [/tex]
[tex] \cos^2 x = \frac{1}{2} [/tex]
[tex] \boxed{\cos x = \pm \frac{\sqrt{2}}{2}} [/tex]
[tex] \boxed{\cos x = \pm \frac{\sqrt{2}}{2} \; \; \sin x = \pm \frac{\sqrt{2}}{2}} [/tex]
[tex] \boxed{x = \pm \frac{\pi}{4} \; \; x = \pm \frac{3 \pi}{4}} [/tex] - solution 1
[tex] \boxed{x = \frac{\pi}{4} \; \; x = \frac{3 \pi}{4} \; \; x = \frac{5 \pi}{4} \; \; x = \frac{7 \pi}{4}} [/tex] - solution 2
:uhh:

By locking on the quadrants, a solution for [itex] x [/itex] can be found for 'special angles', without using inverse functions. However, in this equation, solutions exist in all four quadrants.

A physics professor will accept solution 1 or 2 for [itex] x [/itex], however a mathematics professor will only accept solution 2 for [itex] x [/itex].
 
  • #11
dextercioby
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Nope,as u can see,the initial equations placed no restrictions on the domains of the involved functions.So i'm sorry to dissapoint you,but u have search for solutions in [itex] \mbox{R} [/itex].

Daniel.
 
  • #12
70
0
orion1
u good ,dont get 2 impressed
 
  • #13
70
0
orion i still insist u good ,for the second trig question x=60 or 300 which u got right .see ya
 
  • #14
970
3
Range Rover...


If the range for function 1 is [itex] \left[ 0 . 2 \pi \right] [/itex] then solution 1 and 2 are correct, however if the range for function 1 is [itex](-\infty . +\infty)[/itex] then the number of solutions for [itex]x[/itex] are infinite and the solution for [itex]R[/itex] becomes:

[tex]x = \frac{\pi}{4} + \frac{n \pi}{2} = \frac{\pi}{2} \left( n + \frac{1}{2} \right)[/tex]
[tex]\boxed{ x = \pm \frac{\pi}{2} \left( n + \frac{1}{2} \right)}[/tex]

For function 2, the number of solutions for [itex]x[/itex] are infinite and limited solutions for [itex]x[/itex] becomes:
[tex]x = \cos^{-1} \frac{1}{2}[/tex]
[tex]\boxed{x = \pm \frac{\pi}{3}}[/tex]
[tex]\boxed{x = \frac{\pi}{3} \; \; x = \frac{5 \pi}{3}}[/tex]

For function 2, what is the [itex]x[/itex] solution for [itex]R[/itex]?
 
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