Mt7, Mt19 - Trigonometric equations on given intervals

Mt7, Mt19 - Trigonometric equations on given intervals

Use inverse principal values to find every solution on a stated interval, transform complete arguments, and justify which roots survive identity-based rearrangements.

Principal values and every bounded branch

Why can one inverse-trigonometric key give one answer when a graph shows four intersections? For sin⁡2x=cos⁡x\sin2x=\cos x, one turn of xx 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][-1,1]. Their period is 2π2\pi radians or 360∘360^\circ; tangent's period is π\pi or 180∘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)=π/6\arcsin(1/2)=\pi/6; 5π/65\pi/6 has the same sine but is not that inverse function's output.

The three inverse functions use different ranges: sine uses [−π/2,π/2][-\pi/2,\pi/2], cosine [0,π][0,\pi], tangent (−π/2,π/2)(-\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. Here we list solutions on the given finite intervals; no unrestricted general-solution formula is needed.

Worked example 1: negative sine in degrees

Question. Solve sin⁡θ=−1/2\sin\theta=-1/2 on 0∘≤θ<360∘0^\circ\le\theta<360^\circ.

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

arcsin⁡(−1/2)=−30∘principal valueθ=180∘+30∘, 360∘−30∘quadrants III and IV\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. θ=210∘,330∘\theta=210^\circ,330^\circ.

Check. Both have sine −1/2-1/2; 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⁡θ=−1/2\sin\theta=-1/2 on −180∘≤θ<180∘-180^\circ\le\theta<180^\circ.

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

210∘−360∘=−150∘same sine330∘−360∘=−30∘same sine\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. θ=−150∘,−30∘\theta=-150^\circ,-30^\circ.

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

For tangent, half-turn spacing gives tan⁡θ=−1\tan\theta=-1 the solutions −45∘,135∘-45^\circ,135^\circ on −180∘≤θ<180∘-180^\circ\le\theta<180^\circ. Endpoints also matter: cos⁡x=1\cos x=1 has two inputs on [0,2π][0,2\pi], but only one on [0,2π)[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−π/6)=1/2\sin(2x-\pi/6)=1/2 on 0≤x<2π0\le x<2\pi.

Plan. Solve for u=2x−π/6u=2x-\pi/6, then undo the linear transformation.

−π/6≤u<23π/6transform intervalu=π/6, 5π/6, 13π/6, 17π/6all sine-half branchesx=(u+π/6)/2undo the argument\begin{aligned} -\pi/6&\le u<23\pi/6 &&\text{transform interval}\\ u&=\pi/6,\ 5\pi/6,\ 13\pi/6,\ 17\pi/6 &&\text{all sine-half branches}\\ x&=(u+\pi/6)/2 &&\text{undo the argument} \end{aligned}

Why this step? Solving only for 0≤u<2π0\le u<2\pi would miss half the allowed inputs.

Answer. x=π/6,π/2,7π/6,3π/2x=\pi/6,\pi/2,7\pi/6,3\pi/2.

Check. Substitution gives the four listed arguments, each with sine 1/21/2.

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

Worked example 4: a negative multiple

Question. Solve sin⁡(−2x−π/6)=1/2\sin(-2x-\pi/6)=1/2 on 0≤x<2π0\le x<2\pi.

Plan. Transform both bounds for u=−2x−π/6u=-2x-\pi/6, retaining which endpoint is open.

−25π/6<u≤−π/6reverse the inequalitiesu=−23π/6,−19π/6,−11π/6,−7π/6all allowed branchesx=−(u+π/6)/2recover the variable\begin{aligned} -25\pi/6&<u\le-\pi/6 &&\text{reverse the inequalities}\\ u&=-23\pi/6,-19\pi/6,-11\pi/6,-7\pi/6 &&\text{all allowed branches}\\ x&=-(u+\pi/6)/2 &&\text{recover the variable} \end{aligned}

Why this step? The excluded upper bound for xx becomes the excluded lower bound for uu.

Answer. x=π/2,5π/6,3π/2,11π/6x=\pi/2,5\pi/6,3\pi/2,11\pi/6, in increasing order.

Check. Each angle returns one listed uu-value, and every xx belongs to [0,2π)[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⁡θ\cos\theta is a function value, not yet a solution for θ\theta.

You already know

Use sin⁡2θ=1−cos⁡2θ\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][-1,1].

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

Worked example 5: a quadratic in cosine

Question. Solve 6sin⁡2θ+cos⁡θ−4=06\sin^2\theta+\cos\theta-4=0 on 0∘≤θ<360∘0^\circ\le\theta<360^\circ, to 1 decimal place where necessary.

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

6(1−c2)+c−4=0Pythagorean identity6c2−c−2=(3c−2)(2c+1)=0rearrange and factorc=2/3, −1/2both allowed values\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. θ=48.2∘,120∘,240∘,311.8∘\theta=48.2^\circ,120^\circ,240^\circ,311.8^\circ, with non-exact angles to 1 decimal place.

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

Change only the constant term. One algebraic value now lies outside cosine's range, while the other sits exactly on its boundary.

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