Interval equations and R form

Transform the interval with the angle

An inverse trigonometric function returns a principal value, while an equation can have several solutions in its allowed interval. Keep the whole angle expression and its transformed interval beside the equation. Find every permitted angle there, then translate back to the original variable.

Solve sin 2x = 1/2 for 0° ≤ x < 360°. Put u = 2x, so 0° ≤ u < 720°. Sine is 1/2 at 30° and 150° in the first turn and at the corresponding angles one turn later:

The sine pattern repeats every 360°, and these are the only four angles with sine 1/2 in the transformed interval. All four x values lie in the original interval and satisfy the original equation. Searching only 0° ≤ u < 360° would miss half the requested solutions.

A negative angle coefficient reverses the bounds

For cos(30° − 2x) = 1/2 with 0° ≤ x ≤ 180°, put u = 30° − 2x. At the original endpoints u is 30° and −330°, so the full transformed interval is

The permitted cosine angles are −300° and −60°. Solving 30° − 2x = u gives x = 165° and 45°, respectively. The reversal comes from subtracting 2x: increasing x decreases u. Check the angle equation and the original bounds rather than dividing the interval endpoints as though the coefficient were positive.

Factoring preserves zero alternatives

For 2 sin² x- sin x = 0, factor to

A product is zero when either factor is zero. On 0° ≤ x < 360°, sin x = 0 gives 0° and 180°, while sin x = 1/2 gives 30° and 150°. Thus the solutions are 0°, 30°, 150° and 180°. The upper endpoint 360° is excluded.

Dividing immediately by sin x would lose the first two solutions, because those are exactly the inputs where that division is invalid. If a trigonometric quadratic gives a candidate outside −1 ≤ sin x ≤ 1, reject that candidate before looking for angles. For example, sin x = 2 has no real solution.

Derive R form from the angle-addition identities

A sum of a cosine and a sine of the same angle can be written as one shifted wave. For

expand the right side as R cos x cos α + R sin x sin α. Matching coefficients requires

Choose R > 0. Squaring and adding gives R²(cos²α + sin²α) = 3² + 4², so R = 5. Both coefficients are positive, placing α in quadrant I, with α ≈ 53.13°. The expression ranges from −5 to 5 when x can run through a full turn.

An equivalent sine form is 5 sin(x + β), because expansion requires 5 sin β = 3 and 5 cos β = 4. Here β ≈ 36.87° and α + β = 90°. The two phase angles differ because the chosen sine and cosine forms start at different points of their cycles; either form works when its coefficients and sign are correct.

Check both coefficient signs before choosing the phase

For -3 cos x + 4 sin x = 5 cos(x-α), matching requires cos α = −3/5 and sin α = 4/5. Thus α lies in quadrant II, giving α ≈ 126.87°. A calculator’s principal value for tan⁻¹(−4/3), about −53.13°, has the same tangent but the wrong sine and cosine signs. Expanding with that wrong phase would reconstruct +3 cos x − 4 sin x instead.

Coefficient matching checks a proposed R form directly. It is more reliable than choosing a phase from its tangent alone. If both original coefficients are zero, the expression is the constant zero function, rather than a non-zero wave with a unique phase.

Solve within the shifted interval

To solve 3 cos x + 4 sin x = 2 on 0° ≤ x < 360°, use the positive-coefficient phase above. The equation becomes

Set u = x − α. The permitted interval is −α ≤ u < 360° − α, approximately −53.13° ≤ u < 306.87°. Let γ = cos⁻¹(2/5) ≈ 66.42°. Both γ and 360° − γ ≈ 293.58° lie in that shifted interval; −γ does not. Therefore

Keeping the unrounded phase and inverse value gives x ≈ 119.55° and 346.71°, to two decimal places. Both lie in the original interval and give 2 on substitution before rounding. The shifted interval, rather than the calculator’s one principal angle, establishes which repeated values to retain.

An equation with |right side| > R has no solution. For example, 3 cos x + 4 sin x = 6 would require cosine to be 6/5. At the boundary value R, the wave reaches its maximum rather than crossing that level twice per turn.

Model maxima separately from equation roots

For depth d(t) = 3 + 2 sin(π t/6) metres on 0 ≤ t ≤ 24 hours, the phase is in radians and runs from 0 to 4π. The maximum depth is 5 m, reached at t = 3 h and 15 h. The minimum is 1 m, reached at t = 9 h and 21 h.

Finding when d = 4 m is instead the equation sin(πt/6) = 1/2. Its permitted phase values are π/6, 5π/6, 13π/6 and 17π/6, giving t = 1, 5, 13 and 17 h. A maximum value and the times reaching a chosen value are different answers. The model and its stated time interval, rather than an assumed future repetition, determine which times are relevant.