ONE EIGHTH CONTEST 2025
Presbyterian Boys' Sec. School - 55 points
Oti Boateng SHS - 35 points
T.I. AMASS, Kumasi - 34 points
ONE EIGHTH CONTEST 2025
Our Lady of Grace SHS - 56 points
Bishop Herman College - 39 points
Ejisuman SHS - 27 points
ONE EIGHTH CONTEST 2025
GSTS - 50 points
Nkwatia Presbyterian SHS - 45 points
Ashaiman SHS - 28 points
ROUND 2
TRIGONOMETRY
QUESTION
Given $\cos x=-\dfrac{\sqrt3}{2}$, solve for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{5\pi}{6}$ radians
SOLUTION
$\cos x=-\dfrac{\sqrt3}{2}$
The reference angle is:
$\theta_{\text{ref}}=\dfrac{\pi}{6}$
Cosine is negative in the second quadrant, so:
$x=\pi-\dfrac{\pi}{6}$
$x=\dfrac{5\pi}{6}$
PRACTICE QUESTIONS
1. Solve the equation $\cos x=\dfrac{1}{2}$ for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{\pi}{3}$
SOLUTION
The reference angle is $\theta_{\text{ref}}=\dfrac{\pi}{3}$.
Cosine is positive in the first quadrant, so:
$x=\dfrac{\pi}{3}$
2. Solve the equation $\cos x=\dfrac{\sqrt2}{2}$ for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{\pi}{4}$
SOLUTION
The reference angle is $\theta_{\text{ref}}=\dfrac{\pi}{4}$.
Cosine is positive in the first quadrant, so:
$x=\dfrac{\pi}{4}$
3. Solve the equation $\cos x=\dfrac{\sqrt3}{2}$ for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{\pi}{6}$
SOLUTION
The reference angle is $\theta_{\text{ref}}=\dfrac{\pi}{6}$.
Cosine is positive in the first quadrant, so:
$x=\dfrac{\pi}{6}$
4. Solve the equation $\cos x=-\dfrac{1}{2}$ for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{2\pi}{3}$
SOLUTION
The reference angle is $\theta_{\text{ref}}=\dfrac{\pi}{3}$.
Cosine is negative in the second quadrant, so:
$x=\pi-\dfrac{\pi}{3}$
$x=\dfrac{2\pi}{3}$
5. Solve the equation $\cos x=-\dfrac{\sqrt2}{2}$ for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{3\pi}{4}$
SOLUTION
The reference angle is $\theta_{\text{ref}}=\dfrac{\pi}{4}$.
Cosine is negative in the second quadrant, so:
$x=\pi-\dfrac{\pi}{4}$
$x=\dfrac{3\pi}{4}$
6. Solve the equation $\cos x=-\dfrac{\sqrt3}{2}$ for $x$ in the interval $0\lt x\lt\pi$.
ANSWER: $x=\dfrac{5\pi}{6}$
SOLUTION
The reference angle is $\theta_{\text{ref}}=\dfrac{\pi}{6}$.
Cosine is negative in the second quadrant, so:
$x=\pi-\dfrac{\pi}{6}$
$x=\dfrac{5\pi}{6}$