SL 3.8—Solving trig equations Questionbank

Find the amplitude of $f(x)$.

Determine the possible values of $k$ such that $f(x)$ has a period of $12\pi$.

Given that $k > 0$, find the $x$-coordinates of the maximum and minimum points of $f(x)$ in the domain $0 \le x \le 2\pi$.

Show that the equation may be written as $3\sin x\cos x-\cos x-2=0.$

Factor the left-hand side to show that $\cos x(6\sin x-1)=2.$

Explain why $|\cos x|\le1$ implies $|6\sin x-1|\ge 2$, and hence deduce a quadratic inequality for $\sin x$.

Find all exact values of $\sin x$ consistent with $\cos x(6\sin x-1)=2$, and hence determine all exact values of $x$ in $[0,2\pi)$ that satisfy the original equation.

Sketch the graphs of $y=3\sin(2x)$ and $y=\cos x+2$ for $0\le x\le2\pi$, indicating points of intersection that correspond to your solutions from part 4.

$\sin 2 x=\frac{1}{3}$

$\sin ^{2} x=\frac{1}{3}$

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

$\cos 2 x=\frac{\sqrt{3}}{2}$

State the amplitude and period of $f$.

Determine the $x$-intercepts of $f$ in the given interval, in exact form.

Find the coordinates of the first maximum point of $f$.

Solve, for $0\le x\le2\pi$, the equation $2\sin\left(2x-\frac{\pi}{3}\right)=1.$ Give exact solutions.

Sketch one full period of $y=f(x)$, clearly indicating amplitude, period, and intercepts.

Show that the equation may be written as $4\cos^2x+1-3\sin x=0.$

Express $\cos^2x$ in terms of $\sin x$, and hence obtain a quadratic equation in $\sin x$.

Solve $4\sin^2x+3\sin x-5=0$ exactly for $\sin x$.

Determine all exact values of $x$ in the interval $0\le x<2\pi$ that satisfy the original equation.

Sketch the graphs of $y=2\cos(2x)$ and $y=3\sin x-3$ for $0\le x\le2\pi$, indicating two approximate intersection points corresponding to your solutions from part 4.

Show that the equation may be written as
$4\cos^2x-\sin x-3=0.$

Express $\cos^2x$ in terms of $\sin x$, and hence obtain a quadratic equation in $\sin x$.

Solve $4\sin^2x+\sin x-1=0$ exactly for $\sin x$.

Determine all exact values of $x$ in the given interval that satisfy the original equation.

Sketch the graphs of $y=2\cos(2x)$ and $y=\sin x+1$ for $0\le x\le2\pi$ on the same axes, indicating approximate points of intersection corresponding to your algebraic solutions.

Show that $\log_9(\cos 2y + 2) = \log_3\sqrt{\cos 2y + 2}$

Hence or otherwise solve $\log_3(2\sin y) = \log_9(\cos 2y + 2)$ for $0 < y < \frac{\pi}{2}$.

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

$\sin x=\frac{\sqrt{3}}{2}$

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

$\sin x=-\frac{\sqrt{2}}{2}$

$\sin x=-\frac{\sqrt{3}}{2}$

$\cos x=\frac{1}{2}$, for $0 \leq x \leq 4 \pi$

$\cos 3x=\frac{1}{2}$, for $0 \leq x \leq 2 \pi$