SL 3.8—Solving trig equations Questionbank

For the function $f(x)$ find the amplitude of the function.

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

Find the $x$-coordinates of the minima and maxima of the function $f(x)$ within a period of $12\pi$ and $k>0$ under the domain $x\in[0,2\pi]$

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

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

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

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

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
$2\sin(2x-\tfrac{\pi}{3})=1.$
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 $y=3\sin x-1$ for $0\le x\le2\pi$, indicating 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 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.

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

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

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

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

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

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

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

Show that ${\log_9}\left( {\cos2y + 2} \right) = {\log_3}\sqrt {{\cos2y + 2} }$.

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