by Dr. Carol JVF Burns (website creator)
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• PRACTICE (online exercises and printable worksheets)

that can be transformed into a quadratic equation using an appropriate substitution.

Recall that a quadratic equation in one variable has the form $\,ax^2 + bx + c = 0\,$, for $\,a\ne 0\,$.

For example, ‘$\,2\sin^2 x - \sin x - 1 = 0\,$’ is a pseudo-quadratic equation.
Rewriting as   ‘$\,\color{blue}{2}\color{red}{(\sin x)}^\color{blue}{2} \color{blue}{-} \color{red}{(\sin x)}\color{blue}{ - 1 = 0}\,$’   more clearly illustrates the underlying
‘$\,\color{blue}{2}\color{red}{(\ )}^{\color{blue}{2}} \color{blue}{-} \color{red}{(\ )} \color{blue}{- 1 = 0}\,$’.
The equation ‘$\,2\sin^2 x - \sin x - 1 = 0\,$’ is solved in detail below.

Observe that ‘$\,2\sin^2 x - \sin x - 1 = 0\,$’ is not a true quadratic equation in $\,x\,$,
since such an equation would only be allowed to have $\,x^2\,$, $\,x\,$, and constant terms.
This equation has both ‘$\,\sin x\,$’ and ‘$\,\sin^2 x\,$’ terms.

The equation ‘$\,2\sin^2 x - \sin x - 1 = 0\,$’ can be transformed into a quadratic equation and solved, as follows:

• Use the substitution $\,t = \sin x\,$ to transform to the quadratic equation $\,2t^2 - t - 1 = 0\,$.
That is, replace every ‘$\,\sin x\,$’ with ‘$\,t\,$’.
• Solve the quadratic equation using standard techniques, obtaining $\,t = 1\,$ or $\,t = -\frac 12\,$.
• Use the substitution $\,t = \sin x\,$ a second time to transform back to $\,x\,$.
That is, replace every ‘$\,t\,$’ with ‘$\,\sin x\,$’.
This gives:   $\,\sin x = 1\,$ or $\,\sin x = -\frac 12\,$
Solve using standard techniques.

Here are more examples of pseudo-quadratic equations with their underlying quadratic patterns:
 $x^4 + 2x^2 - 3 = 0$ $(x^2)^2 + 2(x^2) - 3 = 0$ let $\,t = x^2\,$: $t^2 + 2t - 3 = 0$ $x^6 + 3x^3 = 10$ $(x^3)^2 + 3(x^3) = 10$ let $\,t = x^3\,$: $t^2 + 3t = 10$ $\text{e}^{2x} - \text{e}^x - 6 = 0$ $(\text{e}^x)^2 - (\text{e}^x) - 6 = 0$ let $\,t = \text{e}^x\,$: $t^2 - t - 6 = 0$ This equation is solved in detail in Solving Exponential Equations. $2(5x+1)^2 - 7(5x+1) - 9 = 0$ let $\,t = 5x+1\,$: $2t^2 - 7t - 9 = 0$

The equation ‘$\,2\sin^2 x - \sin x - 1 = 0\,$’ is solved in detail in the following example.
The equations in the table above (and many more!) use the same general approach, and are solved in the exercises.

## EXAMPLE: Solving a Pseudo-Quadratic Trigonometric Equation

 $2\sin^2 x - \sin x - 1 = 0$ original equation $2t^2 - t - 1 = 0$ Step 1: Substitution to Transform to Quadratic Equation Recognize the quadratic pattern, and decide on the substitution. Rewrite in the new variable. For this example, we let $\,t = \sin x\,$. $(2t + 1)(t - 1) = 0$ $2t+1 = 0\ \ \text{ or }\ \ t-1 = 0$ $t = -\frac 12\ \ \text{ or }\ \ t = 1$ Step 2: Solve the Resulting Quadratic Equation Write the quadratic equation in standard form, if needed. Solve, using whichever method is easiest: factor and use the zero factor law (Here are examples where the coefficient of the square term is not one.) use the quadratic formula For this example, factoring was easiest. $\sin x = -\frac 12\ \ \text{ or }\ \ \sin x = 1$ Step 3: Transform Back to the Original Variable After this step, you will have (simpler) equation(s) in the original variable. For this example, every occurrence of $\,t\,$ from the end of step 2 is replaced with $\,\sin x\,$. For all integers $\,k\,$, with $\,x\,$ in degrees: solution of ‘$\,\sin x = -\frac 12\,$’ yields: $x = -30^\circ + {360k}^\circ\,$ or $\,x = -150^\circ + {360k}^\circ$ solution of ‘$\,\sin x = 1\,$’ yields: $x = 90^\circ + {360k}^\circ\,$ Step 4: Solve the New (Simpler) Equation(s) Use appropriate techniques. For this example, the new equations are simple trigonometric equations. The solution techniques are presented in Solving Simple Trigonometric Equations.

## EXAMPLE RE-VISITED:Shorten By Leaving Out the Explicit Substitution

A shortened version of the example above appears next.
For people who are comfortable with the process, it's not necessary to explicitly write down the substitution.
If this confuses you, just do it the first way!

 $2\sin^2 x - \sin x - 1 = 0$ original equation $(2\sin x + 1)(\sin x - 1) = 0$ $2\sin x + 1 = 0\ \ \text{ or }\ \ \sin x - 1 = 0$ $\sin x = -\frac 12\ \ \text{ or }\ \ \sin x = 1$ factor and use the zero factor law For all integers $\,k\,$: $x = -\frac{\pi}6 + 2\pi k\,$ or $\,x = -\frac{5\pi}6 + 2\pi k$ or $x = \frac{\pi}2 + 2\pi k$ solve the resulting simple trigonometric equations
Master the ideas from this section