A ‘pseudoquadratic’ (or ‘fake quadratic’) equation is an equation
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 pseudoquadratic 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
quadratic pattern:
‘$\,\color{blue}{2}\color{red}{(\ )}^{\color{blue}{2}} \color{blue}{} \color{red}{(\ )} \color{blue}{ 1 = 0}\,$’.
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:
Here are more examples of pseudoquadratic 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.
$2\sin^2 x  \sin x  1 = 0$  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 $\,t = \sin x\,.$ 
$(2t + 1)(t  1) = 0$ $2t+1 = 0\ \ \text{ or }\ \ t1 = 0$ $t = \frac 12\ \ \text{ or }\ \ t = 1$ 
Step 2: Solve the Resulting Quadratic Equation Write the quadratic equation in standard form, if needed. Solve, using whichever method is easiest:

$\sin x = \frac 12\ \ \text{ or }\ \ \sin x = 1$ 
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. 
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!
Here, the solutions are given in radians, instead of degrees.
$2\sin^2 x  \sin x  1 = 0$  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 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
