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Multiple and shifted angle equations

Solve trigonometric equations containing 2x, 3x or shifted angles. Transform the whole interval, map every root back and preserve denominator restrictions.

Before you startBasic radian trigonometric equations, interval inequalities and angle identities

01 / Solve for the whole angle

A coefficient changes how many periods fit inside the interval.

In sin(2x)=1/2, the input to sine is 2x. As x moves from 0 to 2π, that inner angle moves from 0 to 4π, covering two full sine periods. Stopping after the first pair of sine roots misses half the answers.

Let u=mx+c → transform both interval endpoints → solve in u → use x=(u−c)/m → check the original equation.

Use the model to see every inner-angle root mapped to its corresponding x value. Explanations and practice are independent of animation timing.

Map the entire angle intervalExplore

0 ≤ u < 4π.

Inner roots: π/6, 5π/6, 13π/6, 17π/6.

4 solutions: π/12, 5π/12, 13π/12, 17π/12.

The x interval is always 0≤x<2π. Filled endpoints are included; hollow endpoints are excluded. A negative coefficient reverses the mapping direction.

Equation typeCoefficient of xAdded angle

Watch an expanded angle interval map back

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / A multiple angle expands the search interval

First solve the familiar equation over the larger interval.

Solve sin(2x)=1/2 for 0≤x<2πWorked example

Let u=2x, so 0≤u<4π

There are two sine periods in this interval.

u=π/6, 5π/6, 13π/6, 17π/6

Take both branches in both periods.

Divide every value by 2

x=π/12, 5π/12, 13π/12, 17π/12.

Substitute and check the original interval

Four distinct solutions remain.

Do not divide the right-hand side: sin(2x)=1/2 does not mean sin x=1/4. The 2 is inside the function.

03 / A shift moves both interval endpoints

Undo the shift only after the trigonometric step.

Solve tan(x+π/4)=−1 for 0≤x<2πWorked example

Let u=x+π/4, so π/4≤u<9π/4

Add π/4 to both bounds.

u=3π/4 or 7π/4

Tangent repeats every π.

x=u−π/4

x=π/2 or 3π/2.

Check tangent is defined at both inner angles

Neither u is an odd multiple of π/2.

04 / Combine scaling and shifting

Keep the order of operations explicit.

Solve cos(3x−π/3)=1/2 for 0≤x<2πWorked example

u=3x−π/3 gives −π/3≤u<17π/3

Multiply first, then subtract π/3.

u=−π/3, π/3, 5π/3, 7π/3, 11π/3, 13π/3

The next root 17π/3 is the excluded upper endpoint.

x=(u+π/3)/3

Add π/3 before dividing.

x=0, 2π/9, 2π/3, 8π/9, 4π/3, 14π/9

Six distinct solutions; 0 is allowed.

Writing x=u/3+π/3 would undo the shift incorrectly. The inverse relation is x=u/3+π/9.

05 / A negative coefficient reverses inequalities

Track which endpoint is included after the reversal.

Solve sin(π/3−2x)=1/2 for 0≤x<πWorked example

Let u=π/3−2x

At x=0, u=π/3 is included. As x approaches π, u approaches −5π/3, excluded.

−5π/3

The increasing-order interval has an open left endpoint.

u=−7π/6 or π/6

Both sine values are 1/2.

x=(π/3−u)/2

x=3π/4 or π/12; usually list π/12, 3π/4 in increasing order.

A negative coefficient does not remove solutions. It reverses the order in which their angles are visited.

06 / Treat boundary roots separately

A closed interval can include both ends of a complete period.

Solve cos(2x)=1 for −π≤x≤πWorked example

−2π≤u≤2π

Both endpoints remain included.

u=−2π, 0, 2π

Each is an even multiple of π.

x=−π, 0, π

All three are different allowed inputs.

On the half-open interval −π≤x<π, omit x=π. Always compare with the original x interval after mapping back; a plausible inner angle may correspond to an excluded endpoint.

07 / General solutions avoid a long guess-and-check list

Solve an integer inequality to decide which periods count.

Solve sin(3x)=0 for −π/2≤x≤π/2Worked example

3x=nπ, so x=nπ/3

n is an integer.

−π/2≤nπ/3≤π/2

Multiply by 3/π, which is positive.

−3/2≤n≤3/2

Thus n=−1,0,1.

x=−π/3, 0, π/3

No other integer fits.

For tan(mx+c)=k, you can start from mx+c=arctan k+nπ. For sine and cosine, retain both distinct branches before filtering the integers.

08 / Record excluded inputs before multiplying

An algebraic candidate is not automatically a solution of a fraction.

Consider cos(2x)/(1−sin(2x))=1 for 0≤x<2π. The denominator excludes sin(2x)=1, so x=π/4 and 5π/4 cannot be answers.

Solve while preserving the domainWorked example

cos(2x)+sin(2x)−1=0

Multiplication is valid only at inputs where the denominator is nonzero.

2sin x(cos x−sin x)=0

Use cos(2x)=1−2sin²x and sin(2x)=2sin x cos x.

Candidates: 0, π, π/4, 5π/4

The first factor gives 0,π; the second gives π/4,5π/4.

Reject π/4 and 5π/4

The original fraction is undefined there.

Final answer: x=0 or π

At either input, the original fraction equals 1.

09 / Check the transformed interval and original expression

Use separate checks for completeness and validity.

Completeness: have you covered every period in the transformed interval?
Validity: does every candidate satisfy the original expression and its domain?
Endpoint check: do the inequality signs include or exclude that value?
Inverse check: have you undone addition before division?
Duplicate check: are repeated sine/cosine branches giving the same angle?

Check the range before listing angles.

Worked solution

Cosine would need to equal 3/2. No real solutions.

11 / Recap

Map the interval as carefully as the individual roots.

Use one symbol for the entire trigonometric input.
Transform both bounds, reversing their order for a negative multiplier.
Include every branch and every period in that interval.
Undo a shift before dividing by the multiplier.
Preserve denominator restrictions and original endpoints.
List distinct final x values in increasing order.

Section 1 of 11 · Solve for the whole angle