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cos(2x)+5cos(x)+3=0

cos⁡(2x)+5cos⁡(x)+3=0 \cos (2 x)+5 \cos (x)+3=0

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Full solution

Q. cos⁡(2x)+5cos⁡(x)+3=0 \cos (2 x)+5 \cos (x)+3=0
  1. Rewrite using double angle formula: Rewrite cos⁡(2x)\cos(2x) using the double angle formula: cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1.\newlineCalculation: cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1.
  2. Substitute into original equation: Substitute the expression for cos⁡(2x)\cos(2x) into the original equation.\newlineCalculation: 2cos⁡2(x)−1+5cos⁡(x)+3=02\cos^2(x) - 1 + 5\cos(x) + 3 = 0.
  3. Simplify the equation: Simplify the equation.\newlineCalculation: 2cos⁡2(x)+5cos⁡(x)+2=02\cos^2(x) + 5\cos(x) + 2 = 0.
  4. Factorize the quadratic: Factorize the quadratic equation.\newlineCalculation: \(2\cos(x) + 11)(\cos(x) + 22) = 00.
  5. Solve each factor: Solve each factor for cos⁡(x)\cos(x).\newlineCalculation: 2cos⁡(x)+1=02\cos(x) + 1 = 0 and cos⁡(x)+2=0\cos(x) + 2 = 0.
  6. Solve for cos⁡(x)\cos(x): Solve 2cos⁡(x)+1=02\cos(x) + 1 = 0 for cos⁡(x)\cos(x).\newlineCalculation: cos⁡(x)=−12\cos(x) = -\frac{1}{2}.
  7. Solve for cos⁡(x)\cos(x): Solve cos⁡(x)+2=0\cos(x) + 2 = 0 for cos⁡(x)\cos(x).\newlineCalculation: cos⁡(x)=−2\cos(x) = -2.

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