10 trigonometric equations that separate EXPERTS from beginners

Overview

blackpenredpen works through 10 trigonometric equations on domains ranging from restricted intervals to all real numbers, using angle-addition identities, inverse-trig relationships, graph intersections, and geometric-series sums. The solutions emphasize domain checks and extraneous-root control, including a count of 21 real intersections for cos x = x/(10π) and the rejection of a negative candidate in the final inverse-sine equation.

Key takeaways

The identity a sin x + b cos x can be rewritten as a single shifted sine; recognizing the special-angle coefficients in sin x + √3 cos x produces sin(x + π/3).
For inverse-trigonometric equations, principal ranges and input domains are essential: arcsin x requires −1 ≤ x ≤ 1, and squaring an equation can introduce candidates that fail the original sign condition.
Graph-based counting can reveal solutions that are easy to miss algebraically: cos x = x/(10π) has 21 real solutions, including the endpoint x = 10π but not x = −10π.
Products of cosine terms at doubling angles telescope through sin(2a) = 2 sin a cos a; for cos x cos 2x cos 4x = 1/8, this leads to sin(8x) = sin x, subject to excluding sin x = 0.
A finite-looking sequence 1 + sin²x + sin⁴x + ⋯ is an infinite geometric series, and its sum is valid only when |sin²x| < 1; equating it to 4 yields four solutions on [0, 2π].

Chapters

0:00 Combine Sine and Cosine into One Angle

For sin x + √3 cos x = √2 on [0, 2π], divide by 2 to identify cos(π/3)sin x + sin(π/3)cos x.
Apply the sine addition identity to obtain sin(x + π/3) = √2/2.
The solutions in the requested interval are x = 5π/12 and x = 23π/12.

4:45 Factor a Quadratic in tan x

Treat the equation as a quadratic in tan x and factor it as (tan x − 1)(tan x − √3) = 0.
On [0, 2π], tan x = 1 gives π/4 and 5π/4; tan x = √3 gives π/3 and 4π/3.
The tangent period is π, so each reference-angle solution has a second solution π radians later.

7:41 Turn an Inverse-Sine Product into a Quadratic

For arcsin(x)·arccos(x) = 1/2, first require −1 ≤ x ≤ 1 and use arcsin(x) + arccos(x) = π/2.
Set y = arcsin(x); then y(π/2 − y) = 1/2, a quadratic whose roots are y = (π ± √(π² − 8))/4.
Both angles lie in the principal range of arcsin, giving x = sin((π ± √(π² − 8))/4).

16:36 Count Intersections for cos x = x/(10π)

Compare the cosine curve with the line y = x/(10π); across five positive periods, the curve contributes 10 intersections.
Evenness of cosine gives 10 corresponding intersections on the negative side, but the endpoints are not counted by that interval tally.
At x = 10π, both sides equal 1, adding one more solution; x = −10π is not a solution, so the total is 21.

22:44 Reduce a Triple-Angle Fraction Equation

For cos(3x)/cos x + sin(3x)/sin x = 2, combine fractions using denominator sin x cos x.
The numerator becomes sin(4x), so the left side simplifies to 4 cos(2x), with sin x and cos x nonzero in the original equation.
Solving cos(2x) = 1/2 for x in [0, 2π] gives π/6, 5π/6, 7π/6, and 11π/6.

29:04 Simplify Nested Inverse-Tangent and Sine Functions

Represent tan(arctan x) with a right triangle having opposite side x, adjacent side 1, and hypotenuse √(1 + x²).
This gives cos(arctan x) = 1/√(1 + x²); applying sin(arctan u) = u/√(1 + u²) then reduces the nested expression to 1/√(2 + x²).
Setting that result equal to 1/3 yields x² = 7, so the real solutions are x = ±√7.

33:34 Solve a Trigonometric Base Raised to a Trigonometric Power

For (sin x)^(cos(x − π/4)) = 1 on [0, 2π], consider a base of 1 and a zero exponent as separate cases.
The base-one case gives sin x = 1, or x = π/2; the zero-exponent case gives cos(x − π/4) = 0.
The resulting angles x = 3π/4 and 7π/4 join π/2 as the solutions.

37:49 Use a Doubling Product to Solve cos x cos 2x cos 4x

For cos x cos 2x cos 4x = 1/8, multiply by sin x and repeatedly apply sin(2a) = 2 sin a cos a.
When sin x ≠ 0, the product equation reduces to sin(8x) = sin x.
The two angle families are x = 2kπ/7 and x = (2k + 1)π/9, for integers k; exclude values that are multiples of π because the original product is then ±1, not 1/8.

46:43 Solve a Sum of Two Inverse Sines Without Squaring Too Early

For arcsin x + arcsin(2x) = π/3, first isolate arcsin(2x) and apply sine to both sides.
The sine-difference identity gives 2x = (√3/2)√(1 − x²) − x/2, which simplifies to 5x = √3√(1 − x²).
Squaring gives x² = 3/28, but the unsquared equation requires x ≥ 0; the valid solution is x = √21/14, while the negative root is extraneous.

51:31 Sum an Infinite Geometric Series of Even Sine Powers

Interpret 1 + sin²x + sin⁴x + ⋯ as a geometric series with common ratio sin²x; convergence requires sin²x < 1.
Its sum is 1/(1 − sin²x) = 1/cos²x, so setting the sum equal to 4 gives cos²x = 1/4.
On [0, 2π], the four solutions are π/3, 2π/3, 4π/3, and 5π/3.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, blackpenredpen.