Trigonometric Functions

[christina4444]

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!!!!!

 

[HallsofIvy]

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
\(\displaystyle \frac{1}{sin^2(x)}- \frac{cos^2(x)}{sin^2(x)}= 0\). Then an obvious thing to do is multiply through by \(\displaystyle sin^2(x)\)
to get \(\displaystyle 1- cos^2(x)= 0\).

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

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

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, \(\displaystyle sin^2(x)= 1- 2cos(x)+ cos^2(x)\). Of course, \(\displaystyle sin^2(x)= 1- cos^2(x)\) so we can write this as \(\displaystyle 1- cos^2(x)= 1- 2cos(x)+ cos^2(x)\) which reduces to \(\displaystyle cos^2(x)- 2cos(x)= cos(x)(cos(x)- 2)= 0\). 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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