# Likninger: Oppgaver



:::::::::::::::{exercise} Oppgave 1
:::::::::::::{part} a
Finn de generelle løsningene til likningen

$$
\cos x = \frac{1}{2}
$$


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:::::::::::::{part} b
Finn de generelle løsningene til likningen

$$
\sin x = \frac{1}{2}
$$

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:::::::::::::{part} c
Finn de generelle løsningene til likningen

$$
\cos x = -\dfrac{\sqrt{2}}{2}
$$


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:::::::::::::{part} d
Finn de generelle løsningene til likningen

$$
\sin x = -\dfrac{\sqrt{3}}{2}
$$
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:::::::::::::::{exercise} Oppgave 2

:::::::::::::{part} a
Løs likningen

$$
2\sin x - 1 = 0 \qfor x \in [0, 2\pi\rangle
$$
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:::::::::::::{part} b
Løs likningen

$$
2\cos \left(\pi x\right) = \sqrt{3} \qfor x \in [-4, 4\rangle
$$
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:::::::::::::{part} c
Løs likningen

$$
\tan \left(\dfrac{\pi}{2}x\right) = 1 \qfor x \in [-4, 4\rangle
$$
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:::::::::::::{part} d
Løs likningen

$$
\sin x - \sqrt{3} \cos x = 0 \qfor x \in \langle -\pi, \pi\rangle
$$

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:::::::::::::::{exercise} Oppgave 4
:::::::::::::{part} a
Løs likningen

$$
\left(\sin x - \frac{1}{2}\right)\left(\cos x + \frac{\sqrt{3}}{2}\right) = 0 \qfor x \in [0, 2\pi\rangle
$$


:::{hint} Hint
Hvis to funksjoner $f(x) \cdot g(x) = 0$, så er enten $f(x) = 0$ eller $g(x) = 0$.
:::

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:::::::::::::{part} b
Løs likningen

$$
4 \sin \left(\dfrac{x}{2}\right) + 2 = 0 \qfor x \in [-2\pi, 2\pi\rangle
$$
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:::::::::::::{part} c
Løs likningen

$$
\cos (\pi x) - \sin (\pi x) = 0 \qfor x \in [-2, 2\rangle
$$
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:::::::::::::{part} d
Løs likningen

$$
2\sqrt{3} \sin (2x) + 2 \cos (2x) = 0 \qfor x \in [-2\pi, 2\pi\rangle
$$
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:::::::::::::::{exercise} Oppgave 5
:::::::::::::{part} a
Løs likningen

$$
\cos \left(\dfrac{\pi}{2}x - \dfrac{\pi}{3}\right) = \dfrac{1}{2}
$$
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:::::::::::::{part} b
Løs likningen

$$
\sin \left(\dfrac{\pi}{2}x + \dfrac{\pi}{6}\right) = -\dfrac{\sqrt{2}}{2}
$$
:::::::::::::



:::::::::::::{part} c
Løs likningen

$$
\tan \left(\dfrac{\pi}{2}x - \dfrac{\pi}{4}\right) = 1
$$
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:::::::::::::::{exercise} Oppgave 6
:::::::::::::{part} a
Løs likningen

$$
2 \cos^2 x + \sin x - 2 = 0 \qfor x \in [0, 2\pi\rangle
$$


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:::::::::::::{part} b
Løs likningen

$$
\sin^2 x + \dfrac{1}{2} \sin (2x) = 0 \qfor x \in [0, 2\pi\rangle
$$


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:::::::::::::{part} c
Løs likningen

$$
\sin^2 x - 2 \sin x + 1 = 0 \qfor x \in [0, 2\pi\rangle
$$

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:::::::::::::::{exercise} Oppgave 7
En likning er gitt ved 

$$
\cos^2\left(\dfrac{\pi}{2}x\right) - \sin^2 \left(\dfrac{\pi}{2}x\right) = \dfrac{1}{2}
$$


:::::::::::::{part} a
Finn de generelle løsningene til likningen.


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:::::::::::::{part} b
Bestem $a$ slik at likningen har $6$ løsninger i intervallet $[-a, a\rangle$. 
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:::::::::::::::{exercise} Oppgave 8
:::::::::::::{part} a
Gitt punktene $A(1, 1, 1)$, $B(3, 3, 2)$ og $C(2, 1, 2)$.

Finn vinkelen $\angle BAC$.


:::{hint} Hint
Bruk den geometriske formelen for prikkproduktet 

$$
\vec a \cdot \vec b = \abs{\vec a} \abs{\vec b} \cos \varphi
$$
:::


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