Trigonometric Functions

christina4444

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Oct 25, 2012
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Solve these equations fr exact solutions in the interval [0 degrees, 360 degrees)
csc^2x - cot^2x = 0

2tan^2xsinx - tan^2x = 0

sec^2xtanx = 2tanx

sinx + cosx = 1

I understood my homework until I got to these problems so if someone could please explain these to me that would be greatly appreciated!!!!!
 
Solve these equations fr exact solutions in the interval [0 degrees, 360 degrees)
csc^2x - cot^2x = 0
I personally find "sine" and "cosine" easiest to work with so I would replace cot(x) by cos(x)/sin(x) and csc(x) by 1/sin(x) to get
1sin2(x)cos2(x)sin2(x)=0\displaystyle \frac{1}{sin^2(x)}- \frac{cos^2(x)}{sin^2(x)}= 0. Then an obvious thing to do is multiply through by sin2(x)\displaystyle sin^2(x)
to get 1cos2(x)=0\displaystyle 1- cos^2(x)= 0.

2tan^2xsinx - tan^2x = 0
This obviously factors as tan2(x)(2sin(x)1)=0\displaystyle tan^2(x)(2sin(x)- 1)= 0.

sec^2xtanx = 2tanx
Rewrite as \(\displaystyle sec^2(x)tan(x)- 2tan(x)= tan(x)(sec^2(x)- 2)= 0

sinx + cosx = 1
This can be written as sin(x)= 1- cos(x) and then, squaring both sides, sin2(x)=12cos(x)+cos2(x)\displaystyle sin^2(x)= 1- 2cos(x)+ cos^2(x). Of course, sin2(x)=1cos2(x)\displaystyle sin^2(x)= 1- cos^2(x) so we can write this as 1cos2(x)=12cos(x)+cos2(x)\displaystyle 1- cos^2(x)= 1- 2cos(x)+ cos^2(x) which reduces to cos2(x)2cos(x)=cos(x)(cos(x)2)=0\displaystyle cos^2(x)- 2cos(x)= cos(x)(cos(x)- 2)= 0. Once you have found solutions to that equation, check them in the original equation. Squaring both sides can introduce solutions to the new equation that do not satisfy the original equation.

I understood my homework until I got to these problems so if someone could please explain these to me that would be greatly appreciated!!!!!
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