5.09 Solving equations using radians
Consider the equation \sin \theta = - \dfrac{1}{2}. State the number of solutions for \theta, in the domain \\ 0 \lt \theta \lt \dfrac{\pi}{2}.
For each of the following equations, find the exact value of acute angle x, in the domain \\ 0 \leq x \leq \dfrac{\pi}{2}:
\tan x = 1
\sin x = \dfrac{\sqrt{3}}{2}
\cos x = \dfrac{1}{\sqrt{2}}
2 \tan x = \dfrac{2 \sqrt{3}}{3}
For each of the following, find the exact values of x, where - \pi \lt x \lt \pi:
\tan x = \sqrt{3}
\tan x = \dfrac{1}{\sqrt{3}}
\sin 2 x = \dfrac{1}{2}
\tan \left(x - \dfrac{2 \pi}{3}\right) = 0
For each of the following, find the exact values of \theta, in the given domain:
\tan \theta = 0, where - \pi \leq \theta \leq \pi
\sin \theta = 0, where - 4 \pi \leq \theta \leq 4 \pi
12 \cos \theta - 6 \sqrt{3} = 0, where - \pi \lt \theta \lt \pi
4 \sin \theta - 3 = - 5, where - 2\pi \leq \theta \leq 2\pi
\cos ^{2}\left(\dfrac{\theta}{2}\right) - 1 = 0, where - 2\pi \leq \theta \leq 2\pi
4 \sin ^{2}\left(\theta\right) = 1, where - 4 \pi \lt \theta \lt 4 \pi
For each of the following, find the exact values of x, over the domain \left[ 0 , 2 \pi \right):
\sin x = \dfrac{1}{2}
\cos x = - \dfrac{\sqrt{3}}{2}
6 \cos x + 2 = - 1
6 \cos x - 3 \sqrt{2} = 0
2 \tan x + 3 = 5
2 \sin x + 2 = 1
\cos x \tan x = \cos x
\tan ^{2}\left(x\right) + 2 \tan x + 1 = 0
2 \sin 3 x - \sqrt{2} = 0
\sin \left(\dfrac{x}{2}\right) = \dfrac{\sqrt{3}}{2}
2 \sin \left(\dfrac{x}{2}\right) = \sqrt{3}
\sin \left(\dfrac{x}{2}\right) = - \cos \left(\dfrac{x}{2}\right)
\cos \left(\dfrac{x}{2}\right) = 1 - \cos \left(\dfrac{x}{2}\right)
\sin \left(\dfrac{x}{2}\right) = 1 - \sin \left(\dfrac{x}{2}\right)
\cos ^{2}\left(\dfrac{x}{2}\right) - 1 = 0
\sin ^{2}\left(\dfrac{x}{2}\right) - 1 = 0
36 \left(1 - \cos x\right) \left(1 + \cos x\right) = 27
For each of the following, find the exact values of x, over the domain 0 \leq x \leq 2 \pi:
\sin x = 1
\cos x = - 1
\sin x = \dfrac{1}{2}
\sin x = - \dfrac{1}{\sqrt{2}}
\cos x = - \dfrac{1}{2}
\tan x = - \dfrac{1}{\sqrt{3}}
10 \sin x - 5 \sqrt{3} = 0
4 \cos x - 2 = 0
2 \sqrt{3} \tan x - 2 = 0
\sin 2 x = \dfrac{1}{\sqrt{2}}
\tan 3 x = - \dfrac{1}{\sqrt{3}}
\sin \left(x + \dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{2}}
\cos \left(x + \dfrac{\pi}{6}\right) = - \dfrac{1}{2}
\sin x = - \cos x
\sin \left(x + \dfrac{\pi}{3}\right) = \dfrac{1}{2}
\sqrt{3} \tan \left(\dfrac{x}{2}\right) = - 3
\cos \left(x + \dfrac{\pi}{4}\right) = - \dfrac{1}{\sqrt{2}}
\sin \left(x - \dfrac{4 \pi}{3}\right) = \dfrac{1}{2}
Find the exact solutions of the following equations where 0 \leq \theta \leq 2 \pi:
Consider the equation \cos \theta = 0.42. State the number of solutions for \theta over the domain \\ 0 \lt \theta \lt 2\pi.
For each of the following equations:
Find the acute angle \theta in radians that solves the equation, correct to two decimal places.
\sin \theta = 0.3420
\cos \theta = 0.9063
\tan \theta = 0.8340
\cos \theta = 0.2345
\sin \theta = 0.9921
\tan \theta = 0.7743
\cos \theta = 0.67
\sin \theta = 0.091
\cos 2 \theta = 0.9
Jessica finds the acute angle solution to the equation \sin x = k, is x = p.
Find another solution for the equation over the domain 0 \lt x \lt 2\pi, in terms of p.
Toby finds the acute angle solution to the equation \cos x = m, is x = n.
Find another solution for the equation over the domain 0 \lt x \lt 2\pi, in terms of n.
Sarah finds the acute angle solution to the equation \tan x = h, is x = t.
Find another solution for the equation over the domain 0 \lt x \lt 2\pi, in terms of t.