1.05o - Trigonometric equations on given intervals | OCR A-Level Mathematics A

Find every permitted branch in a degree or radian interval, solve equations with transformed arguments and quadratic functions, and use identities without losing or adding roots.

Principal values and every bounded branch

Why can one inverse-trigonometric key give one answer when a graph shows four intersections? For $\sin2x=\cos x$ , one turn of $x$ contains four solutions. We will find all four, then explain what a calculator's single principal value has left out.

You already know

Sine and cosine lie in $[-1,1]$ . Their period is $2\pi$ radians or $360^\circ$ ; tangent's period is $\pi$ or $180^\circ$ , with undefined inputs at odd right angles.

The inverse key chooses one representative by restricting the possible outputs.

Principal value

A principal value is the single output selected by an inverse trigonometric function's restricted range. For example, $\arcsin(1/2)=\pi/6$ ; $5\pi/6$ has the same sine but is not that inverse function's output.

The three inverse functions use different ranges: sine uses $[-\pi/2,\pi/2]$ , cosine $[0,\pi]$ , tangent $(-\pi/2,\pi/2)$ . The equation asks for every permitted input, rather than just that principal representative. These inverse ranges are not in the formula sheet: learn them.

Worked example 1: negative sine in degrees

Question. Solve $\sin\theta=-1/2$ on $0^\circ\le\theta<360^\circ$ .

Plan. Find the reference angle and use the quadrants where sine is negative.

\begin{aligned} \arcsin(-1/2)&=-30^\circ &&\text{principal value}\\ \theta&=180^\circ+30^\circ,\ 360^\circ-30^\circ &&\text{quadrants III and IV} \end{aligned}

Why this step? The principal value is outside the interval, but two angles inside have the same sine.

Answer. $\theta=210^\circ,330^\circ$ .

Check. Both have sine $-1/2$ ; the endpoints have sine zero.

Change only the interval to a turn centred on zero.

Worked example 2: the same sine, a different interval

Question. Solve $\sin\theta=-1/2$ on $-180^\circ\le\theta<180^\circ$ .

Plan. Translate the previous solutions by a full turn where necessary.

\begin{aligned} 210^\circ-360^\circ&=-150^\circ &&\text{same sine}\\ 330^\circ-360^\circ&=-30^\circ &&\text{same sine} \end{aligned}

Why this step? Periodicity preserves the equation, while the new interval selects different representatives.

Answer. $\theta=-150^\circ,-30^\circ$ .

Check. Both lie in the new interval and both have negative sine.

For tangent, half-turn spacing gives $\tan\theta=-1$ the solutions $-45^\circ,135^\circ$ on $-180^\circ\le\theta<180^\circ$ . Endpoints also matter: $\cos x=1$ has two inputs on $[0,2\pi]$ , but only one on $[0,2\pi)$ . At an extremum, reflected branches coincide; do not count the same input twice.

Multiples and shifted arguments

A multiple argument travels through more than one wave while the original variable travels through one turn. Transform the whole interval first.

Worked example 3: a shifted double argument

Question. Solve $\sin(2x-\pi/3)=1/2$ on $0\le x<2\pi$ .

Plan. Solve for $u=2x-\pi/3$ , then undo the linear transformation.

\begin{aligned} -\pi/3&\le u<11\pi/3 &&\text{transform interval}\\ u&=\pi/6,\ 5\pi/6,\ 13\pi/6,\ 17\pi/6 &&\text{all sine-half branches}\\ x&=(u+\pi/3)/2 &&\text{undo the argument} \end{aligned}

Why this step? Solving only for $0\le u<2\pi$ would miss half the allowed inputs.

Answer. $x=\pi/4,7\pi/12,5\pi/4,19\pi/12$ .

Check. Substitution gives the four listed arguments, each with sine $1/2$ .

Change only the sign of the coefficient of $x$ . Inequalities reverse when multiplied by a negative number.

Worked example 4: a negative multiple

Question. Solve $\sin(-2x-\pi/3)=1/2$ on $0\le x<2\pi$ .

Plan. Transform both bounds for $u=-2x-\pi/3$ , retaining which endpoint is open.

\begin{aligned} -13\pi/3&<u\le-\pi/3 &&\text{reverse the inequalities}\\ u&=-23\pi/6,-19\pi/6,-11\pi/6,-7\pi/6 &&\text{all allowed branches}\\ x&=-(u+\pi/3)/2 &&\text{recover the variable} \end{aligned}

Why this step? The excluded upper bound for $x$ becomes the excluded lower bound for $u$ .

Answer. $x=5\pi/12,3\pi/4,17\pi/12,7\pi/4$ , in increasing order.

Check. Each angle returns one listed $u$ -value, and every $x$ belongs to $[0,2\pi)$ .

Tangent works in the same way, with branches separated by a half-turn in its argument. Its undefined inputs remain excluded.

Quadratic equations in one function

Solve the algebraic equation first, then recover the angles. A root for $\cos\theta$ is a function value, not yet a solution for $\theta$ .

You already know

Use $\sin^2\theta=1-\cos^2\theta$ to leave one function. Factor a quadratic or use the quadratic formula. Reject sine or cosine values outside $[-1,1]$ .

The substitution $c=\cos\theta$ keeps the algebra readable while you solve it.

Worked example 5: a quadratic in cosine

Question. Solve $6\sin^2\theta+\cos\theta-4=0$ on $0^\circ\le\theta<360^\circ$ , to 1 decimal place where necessary.

Plan. Replace sine squared, factor in cosine, then find each angle branch.

\begin{aligned} 6(1-c^2)+c-4&=0 &&\text{Pythagorean identity}\\ 6c^2-c-2&=(3c-2)(2c+1)=0 &&\text{rearrange and factor}\\ c&=2/3,\ -1/2 &&\text{both allowed values} \end{aligned}

Why this step? One cosine value usually corresponds to two different angles in a full turn.

Answer. $\theta=48.2^\circ,120^\circ,240^\circ,311.8^\circ$ , with non-exact angles to 1 decimal place.

Check. Each value of $c$ satisfies the quadratic; substitute unrounded inverse-cosine values before rounding.

Change only the constant term. Both algebraic values will now be impossible.

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1.05o - Trigonometric equations on given intervals

Principal values and every bounded branch

Multiples and shifted arguments

Quadratic equations in one function

Continue this lesson.

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