Quadratic Trigonometric Equations (DP IB Analysis & Approaches (AA): SL): Revision Note

Quadratic trigonometric equations

How do I solve quadratic trigonometric equations involving a single ratio?

• STEP 1
  Make a substitution equal to the trig ratio

  • e.g. for $2{\cos }^{2}x+3\cos x-2=0$

    • let $y=\cos x$

    • the equation becomes $2y^2+3y-2=0$

• STEP 2
  Solve the new quadratic equation

  • You can factorise, complete the square, use the formula or use your GDC

    • $\left(2y-1\right)\left(y+2\right)=0$

    • $y=\frac{1}{2}$ or $y=-2$

• STEP 3
  Replace the variable with the substitution and reject any that do not have solutions

  • Remember the values of sin and cos range from -1 to 1

    • $\cos x=\frac{1}{2}$

    • $\cos x=-2$ has no solutions as $-2<-1$

• STEP 4
  Solve the trig equation(s) for the given interval

How do I solve trigonometric equations involving multiple ratios?

If your equation involves tan and another ratio

• Use $\tan x=\frac{\sin x}{\cos x}$ to write the equation in terms of sin and cos only

  • e.g. $2\sin x=\tan x$ becomes $2\sin x=\frac{\sin x}{\cos x}$

• Get rid of any fractions

  • e.g. $2\sin x\cos x=\sin x$

• Make the equation equal to zero

  • $2\sin x\cos x-\sin x=0$

• Factorise the expression

  • e.g. $\sin x\left(2\cos x-1\right)=0$

• Set each factor equal to zero and solve

  • e.g. $\sin x=0$ and $\cos x=\frac{1}{2}$

If your equation involves sin² and/or cos²

• Use ${\sin }^{2}x+{\cos }^{2}x=1$ to write the equation in terms of only one ratio

  • e.g. $3{\cos }^{2}x+4\sin x=4$ can be written as $3\left(1-{\sin }^{2}x\right)+4\sin x=4$

• Make the equation equal to zero

  • e.g. $3-3{\sin }^{2}x+4\sin x=4$ becomes $3{\sin }^{2}x-4\sin x+1=0$

• Solve the quadratic trig equation

  • e.g. $\sin x=\frac{1}{3}$ and $\sin x=1$