solve this hint of trignomatry.
Answers
Answer:
Solve sin(x) + 2 = 3 over the interval 0° ≤ x < 360°
Just as with linear equations, I'll first isolate the variable-containing term:
sin(x) + 2 = 3
sin(x) = 1
Now I'll use the reference angles I've memorized to get my final answer.
Note: The instructions gave me the interval in terms of degrees, which means that I'm supposed to give my answer in degrees. Yes, the sine, on the first period, takes on the value of 1 at \small{ \frac{\pi}{2} }
2
π
radians, but that's not the angle-measure type they're wanting, and using this as my answer would probably result in my at least losing a few points on this question.
So, in degrees, my answer is:
x = 90°
Solve tan2(θ) + 3 = 0 on the interval 0° ≤ θ < 360°
There's the temptation to quickly recall that the tangent of 60° involves the square root of 3 and slap down an answer, but this equation doesn't actually have a solution. I can see this when I slow down and do the steps. My first step is:
tan2(θ) = –3
Can any square (of a tangent, or of any other trig function) be negative? No! So my answer is:
no solution
Solve \mathbf{\color{green}{\small{ 2 \cos^2(\mathit{x}) - \sqrt{3\,} \cos(\mathit{x}) = 0 }}}2cos
2
(x)−
3
cos(x)=0 on the interval 0° ≤ x < 360°
The left-hand side of this equation factors. I'm used to doing simple factoring like this:
2y2 + 3y = 0
y (2y + 3) = 0
...and then solving each of the factors. The same sort of thing works here. To solve the equation they've given me, I will start with the factoring:
\small{ 2 \cos^2(x) - \sqrt{3\,} \cos(x) = 0 }2cos
2
(x)−
3
cos(x)=0
\small{ \cos(x)\,\left(2 \cos(x) - \sqrt{3\,}\right) = 0 }cos(x)(2cos(x)−
3
)=0
\small{ \cos(x) = 0\quad\mathsf{ or }\quad 2 \cos(x) - \sqrt{3\,} = 0 }cos(x)=0or2cos(x)−
3
=0
\small{ \cos(x) = 0\quad\mathsf{ or }\quad \cos(x) = \dfrac{\sqrt{3\,}}{2} }cos(x)=0orcos(x)=
2
3