- #1

- 3

- 1

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

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- Thread starter needhelp123
- Start date

- #1

- 3

- 1

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

- #2

- 13,114

- 666

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

- 3

- 1

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

| (line) is absolute value

| (line) is absolute value

- #4

- 13,114

- 666

[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

Science Advisor

- 2,788

- 13

needhelp123 said:

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

- 970

- 3

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]

[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]

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- #7

- 13,114

- 666

U have to disregard [itex] \cos x=2 [/itex],okay?

Daniel.

Daniel.

- #8

Zurtex

Science Advisor

Homework Helper

- 1,120

- 1

Unless the question asks for complex solutions.dextercioby said:U have to disregard [itex] \cos x=2 [/itex],okay?

Daniel.

- #9

- 13,114

- 666

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

Daniel.

- #10

- 970

- 3

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

- 13,114

- 666

Daniel.

- #12

- 70

- 0

orion1

u good ,dont get 2 impressed

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

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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