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Difficulty: Very hardTrigonometric Graphs and Simple Equations

Find the total number of distinct solutions to the trigonometric equation 2cos⁡2(2x)+sin⁡(2x)−1=02\cos^2(2x) + \sin(2x) - 1 = 0 in the interval 0∘≤x≤360∘0^\circ \le x \le 360^\circ.

Answer: 6

Answer

The total number of distinct solutions is 6.
Substituting cos⁡2(2x)=1−sin⁡2(2x)\cos^2(2x) = 1 - \sin^2(2x) gives the quadratic 2sin⁡2(2x)−sin⁡(2x)−1=02\sin^2(2x) - \sin(2x) - 1 = 0, which factors into (2sin⁡(2x)+1)(sin⁡(2x)−1)=0(2\sin(2x) + 1)(\sin(2x) - 1) = 0. For 0∘≤x≤360∘0^\circ \le x \le 360^\circ, the angle argument 2x2x covers 0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ. The equation sin⁡(2x)=1\sin(2x) = 1 provides 2 values for xx (45∘,225∘45^\circ, 225^\circ), while sin⁡(2x)=−12\sin(2x) = -\frac{1}{2} provides 4 values for xx (105∘,165∘,285∘,345∘105^\circ, 165^\circ, 285^\circ, 345^\circ). Summing these gives 6 distinct solutions in total.

Step-by-Step Solution

1
Use the Pythagorean trigonometric identity cos⁡2(2x)=1−sin⁡2(2x)\cos^2(2x) = 1 - \sin^2(2x) to express the entire equation in terms of sin⁡(2x)\sin(2x).
2(1−sin⁡2(2x))+sin⁡(2x)−1=0  ⟹  2−2sin⁡2(2x)+sin⁡(2x)−1=02(1 - \sin^2(2x)) + \sin(2x) - 1 = 0 \implies 2 - 2\sin^2(2x) + \sin(2x) - 1 = 0
Converting all trigonometric terms to a single function allows the equation to be solved as a polynomial.
2
Rearrange and factorize the resulting quadratic equation in terms of sin⁡(2x)\sin(2x).
2sin⁡2(2x)−sin⁡(2x)−1=0  ⟹  (2sin⁡(2x)+1)(sin⁡(2x)−1)=02\sin^2(2x) - \sin(2x) - 1 = 0 \implies (2\sin(2x) + 1)(\sin(2x) - 1) = 0
Factorization splits the quadratic trigonometric equation into two simple linear trigonometric equations.
3
Determine the expanded domain for 2x2x given 0∘≤x≤360∘0^\circ \le x \le 360^\circ.
0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ
Multiplying the bounds of xx by 2 accounts for two full rotations in the unit circle.
4
Solve the first linear equation sin⁡(2x)=1\sin(2x) = 1 within 0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ.
2x=90∘,450∘  ⟹  x=45∘,225∘2x = 90^\circ, 450^\circ \implies x = 45^\circ, 225^\circ (2 distinct solutions)
The sine function equals 1 at 90∘90^\circ in the first revolution and at 90∘+360∘=450∘90^\circ + 360^\circ = 450^\circ in the second revolution.
5
Solve the second linear equation sin⁡(2x)=−12\sin(2x) = -\frac{1}{2} within 0∘≤2x≤720∘0^\circ \le 2x \le 720^\circ.
2x=210∘,330∘,570∘,690∘  ⟹  x=105∘,165∘,285∘,345∘2x = 210^\circ, 330^\circ, 570^\circ, 690^\circ \implies x = 105^\circ, 165^\circ, 285^\circ, 345^\circ (4 distinct solutions)
The sine function is negative in the 3rd and 4th quadrants of both revolutions.
6
Combine the solution counts from both cases.
Total number of solutions = 2+4=62 + 4 = 6.
Adding the valid solutions from both factor equations gives the complete set of roots.

Key Concept

Solving quadratic trigonometric equations across multiple revolutions
Estimated Time:3m 0s
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Topics

Trigonometric Graphs and Simple Equations

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