Question

$$\begin{array}{cc}TrigonometricEquations& GeneralSolutions\\ 1.7co{s}^{2}x+si{n}^{2}x=4& P.n\pi \pm \frac{\pi }{4},wheren\in I\\ 2.si{n}^{2}x=\frac{1}{2}& Q.n\pi \pm \frac{\pi }{3},wheren\in I\\ 3.ta{n}^{2}x+3=0& R.n\pi \pm \frac{\pi }{6},wheren\in I\\ 4.3ta{n}^{2}x-1=0& S.Nosolution\end{array}$$

• A. 1 - Q, 2 - P, 3 - S, 4 - R
• B. 1 - Q, 2 - P, 3 - Q, 4 - R
• C. 1 - R, 2 - R, 3 - Q, 4 - R
• D. 1 - R, 2 - P, 3 - S, 4 - P

Answer 1

The correct option is A

1 - Q, 2 - P, 3 - S, 4 - R

7 $co{s}^2$ x + 3 $si{n}^2$ x = 4

We know, identity $si{n}^2$ x + $co{s}^2$ x = 1

(4 + 3) $co{s}^2$ x + 3 $si{n}^2$ x = 4

4 $co{s}^2$ x + 3 $co{s}^2$ + 3 $si{n}^2$ x = 4

4 $co{s}^2$x + 3($si{n}^2$ x + $co{s}^2$ x) = 4

4 $co{s}^2$ x + 3 = 4

4 $co{s}^2$ x = 1

$co{s}^2$ x = 1/4

$co{s}^2$ x = ${\left(\frac{1}{2}\right)}^2$ = $co{s}^2$ $\frac{\pi }{3}$

General solutions is x = nπ±$\frac{\pi }{3}$ , Where n ∈ Z

2 . $si{n}^2$ x = $\frac{1}{2}$

$si{n}^2$ x = ${\left(\frac{1}{\sqrt{3}}\right)}^2$ = $si{n}^2$ $\frac{\pi }{4}$

General solution = nπ ± $\frac{\pi }{4}$

3. $ta{n}^2$ x + 3 = 0

$ta{n}^2$ x =-3

square of trigonometric function cannot be negative .

no solution

4. 3$ta{n}^2$ x - 1=0

$ta{n}^2$ x = $\frac{1}{3}$

$ta{n}^2$ x = ${\left(\frac{1}{\sqrt{3}}\right)}^2$

$ta{n}^2$ x = $ta{n}^2$ ($\frac{\pi }{6}$)

So, general solution is x = nπ± $\frac{\pi }{6}$