Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If the equation <math>\cos 3x \cos^3 x + \sin 3x \sin^3 x = 0</math>, then <math>x</math> is equal to</p>
<p>(a) <math>(2n+1)\frac{\pi}{4}</math></p>
<p>(b) <math>(2n-1)\frac{\pi}{4}</math></p>
<p>(c) <math>n\frac{\pi}{2}</math></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the triple angle formulas cos 3x = 4cos³x - 3cos x and sin 3x = 3sin x - 4sin³x to rewrite the equation, then factor and solve the resulting trigonometric equation.
<p><strong>Step 1:</strong> Start with the given equation: cos 3x cos³ x + sin 3x sin³ x = 0</p><p><strong>Step 2:</strong> Apply triple angle formulas: cos 3x = 4cos³ x - 3cos x and sin 3x = 3sin x - 4sin³ x</p><p><strong>Step 3:</strong> Substitute these formulas:<br/>(4cos³ x - 3cos x)cos³ x + (3sin x - 4sin³ x)sin³ x = 0</p><p><strong>Step 4:</strong> Expand:<br/>4cos⁶ x - 3cos⁴ x + 3sin⁴ x - 4sin⁶ x = 0</p><p><strong>Step 5:</strong> Rearrange and factor. Notice that:<br/>4(cos⁶ x - sin⁶ x) - 3(cos⁴ x - sin⁴ x) = 0</p><p><strong>Step 6:</strong> Factor using difference of cubes and difference of squares:<br/>4(cos² x - sin² x)(cos⁴ x + cos² x sin² x + sin⁴ x) - 3(cos² x - sin² x)(cos² x + sin² x) = 0<br/>(cos² x - sin² x)[4(cos⁴ x + cos² x sin² x + sin⁴ x) - 3] = 0</p><p><strong>Step 7:</strong> Since cos⁴ x + sin⁴ x = 1 - 2cos² x sin² x, simplify to get:<br/>(cos² x - sin² x)[4(1 - cos² x sin² x) - 3] = 0<br/>cos 2x · cos 2x = 0</p><p><strong>Step 8:</strong> This gives cos 2x = 0</p><p><strong>Step 9:</strong> Solve cos 2x = 0:<br/>2x = (2n+1)π/2<br/>x = (2n+1)π/4</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A