Question and Answer (Board Exam)

1. (CBSE 2018) Relations: Principal Solutions

Question: Find the principal solutions of the equation $\tan x = \sqrt{3}$.

Explanation: Principal solutions are those in the interval $[0, 2\pi)$.

1. Identify Quadrants: $\tan x$ is positive in Quadrant I and Quadrant III.

2. Find Q1 Solution: The base angle $\alpha$ where $\tan \alpha = \sqrt{3}$ is $\alpha = \pi/3$.

3. Find Q3 Solution: The rule for Q3 is $\pi + \alpha$. So, $x = \pi + \pi/3 = 4\pi/3$.

Answer: The principal solutions are $\pi/3$ and $4\pi/3$.

2. (CBSE 2019) Relations: Principal Solutions

Question: Find the principal solutions of $\sec x = 2$.

Explanation:

1. Convert to Cosine: The equation $\sec x = 2$ is equivalent to $\cos x = 1/2$.

2. Identify Quadrants: $\cos x$ is positive in Q1 and Q4.

3. Find Q1 Solution: The angle is $x = \pi/3$.

4. Find Q4 Solution: The rule is $2\pi - \alpha$. So, $x = 2\pi - \pi/3 = 5\pi/3$.

Answer: The principal solutions are $\pi/3$ and $5\pi/3$.

3. (CBSE 2020) Relations: General Solution

Question: Find the general solution of the equation $\cos(4x) = \cos(2x)$.

Explanation: We use the general solution for $\cos A = \cos B$, which is $A = 2n\pi \pm B$.

Here, $A=4x, B=2x$. So, $4x = 2n\pi \pm 2x$.

• Case 1 (+):

  $4x = 2n\pi + 2x$

  $\implies 2x = 2n\pi \implies x=n\pi$.

• Case 2 (-):

  $4x = 2n\pi - 2x$

  $\implies 6x = 2n\pi \implies x=n\pi/3$.

The set of solutions $\{n\pi\}$ is a subset of $\{n\pi/3\}$. The complete general solution is the larger set.

Answer: The general solution is $x=n\pi/3, n \in \mathbb{Z}$.

4. (CBSE 2017) Relations: General Solution

Question: Find the general solution of the equation $\sin x = -\frac{\sqrt{3}}{2}$.

Explanation:

1. Find a Principal Value: We need an angle $\alpha$ such that $\sin\alpha = -\sqrt{3}/2$. We can choose the angle in Q4, $\alpha = -\pi/3$.

2. Apply General Formula: The solution for $\sin x = \sin\alpha$ is $$x = n\pi + (-1)^n \alpha$$

3. Substitute: $$x = n\pi + (-1)^n(-\pi/3)$$ $$= n\pi - (-1)^n(\pi/3)$$

Answer: $x = n\pi - (-1)^n (\pi/3), n \in \mathbb{Z}$.

5. (CBSE 2022) Relations: General Solution

Question: Find the general solution of the equation $\tan(3x) = -1$.

Explanation:

1. Find Principal Value: We know $\tan(\pi/4)=1$. Since tangent is an odd function, $\tan(-\pi/4)=-1$. So, we can write $\tan(3x) = \tan(-\pi/4)$.

2. Apply General Formula: For $\tan A = \tan B$, the solution is $A = n\pi + B$.

   $3x = n\pi - \pi/4$.

3. Solve for x: Divide by 3.

Answer: $$x = \frac{n\pi}{3} - \frac{\pi}{12}, n \in \mathbb{Z}$$

6. (CBSE Sample Paper) Relations: Squared Functions

Question: Find the general solution of $\sin^2(2x) = 1/4$.

Explanation:

1. Find Principal Value: $\sin^2(2x) = (1/2)^2$. We know $\sin(\pi/6) = 1/2$. So, the equation is $$\sin^2(2x) = \sin^2(\pi/6)$$

2. Apply General Formula: The solution for $\sin^2A = \sin^2\alpha$ is $$A = n\pi \pm \alpha$$

   Here, $A=2x, \alpha=\pi/6$. So, $2x = n\pi \pm \pi/6$.

3. Solve for x: Divide by 2.

Answer: $$x = \frac{n\pi}{2} \pm \frac{\pi}{12}, n \in \mathbb{Z}$$

7. (CBSE 2018) Equations Reducible to Simpler Forms

Question: Solve the equation $2\cos^2x + 3\sin x = 0$.

Explanation:

1. Convert to a single function: Use $\cos^2x = 1-\sin^2x$. The equation becomes $$2(1-\sin^2x) + 3\sin x = 0$$

2. Form a quadratic: $$2-2\sin^2x+3\sin x=0$$ $$\implies 2\sin^2x - 3\sin x - 2 = 0$$

3. Solve the quadratic: Let $y=\sin x$. We have $$2y^2-3y-2=0$$, which factors as $(2y+1)(y-2)=0$. The solutions are $y=-1/2$ and $y=2$.

4. Find valid solutions: This means $\sin x = -1/2$ or $\sin x = 2$. Since the range of sine is $[-1, 1]$, $\sin x = 2$ is impossible.

5. Find the general solution: For $\sin x = -1/2 = \sin(-\pi/6)$, the general solution is $$x = n\pi + (-1)^{n+1} (\pi/6)$$

Answer: $$x = n\pi + (-1)^{n+1} (\pi/6), n \in \mathbb{Z}$$.

8. (CBSE 2021) Equations Reducible to Simpler Forms

Question: Find the number of solutions of the equation $2\sin^2x + 5\sin x - 3 = 0$ in the interval $[0, 3\pi]$.

Explanation:

1. Solve the Quadratic: Let $y=\sin x$. The equation is $2y^2+5y-3=0$, which factors to $(2y-1)(y+3)=0$. This gives $\sin x = 1/2$ or $\sin x = -3$. The second case is impossible.

2. Find Solutions for $\sin x = 1/2$: We need to find all angles in $[0, 3\pi]$ where the sine is 1/2.

   • In $[0, 2\pi]$: Solutions are in Q1 and Q2, which are $\pi/6$ and $5\pi/6$.

   • In $[2\pi, 4\pi]$: The next solutions are $2\pi+\pi/6 = 13\pi/6$ and $2\pi+5\pi/6 = 17\pi/6$.

3. Check Interval: The interval is $[0, 3\pi]$. Since $3\pi = 18\pi/6$, all four solutions found ($\pi/6, 5\pi/6, 13\pi/6, 17\pi/6$) are within this interval.

Answer: There are 4 solutions.

9. (CBSE 2019) Equations Reducible to Simpler Forms

Question: Solve $\sin x + \sin(3x) + \sin(5x) = 0$.

Explanation:

1. Group and apply sum-to-product: Group $(\sin(5x)+\sin x)$. Use $$\sin A + \sin B = 2\sin((A+B)/2)\cos((A-B)/2)$$ $$2\sin(3x)\cos(2x) + \sin(3x) = 0$$ $$\implies \sin(3x)(2\cos(2x)+1)=0$$
2. Solve the two cases:
   • Case 1: $$\sin(3x)=0 \implies 3x=n\pi \implies x=n\pi/3$$
   • Case 2: $$2\cos(2x)+1=0 \implies \cos(2x)=-1/2$$ The general solution is $$2x=2n\pi \pm 2\pi/3 \implies x=n\pi \pm \pi/3$$

Answer: The solutions are $x=n\pi/3$ and $x=n\pi \pm \pi/3$.

10. (CBSE 2016) Equations Reducible to Simpler Forms

Question: Solve the equation $\tan x + \cot x = 2$.

Explanation:

1. Convert to Sine and Cosine: $$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 2$$

2. Simplify: $$\frac{\sin^2x + \cos^2x}{\sin x \cos x} = 2 \implies \frac{1}{\sin x \cos x} = 2$$

3. Use Double Angle Identity: Rearrange to $1=2\sin x \cos x$, which is $\sin(2x) = 1$.

4. Find General Solution: The general solution for $\sin\theta=1$ is $$\theta = 2n\pi + \pi/2$$ $$2x = 2n\pi + \pi/2$$ Dividing by 2 gives the final answer.

Answer: $$x = n\pi + \pi/4$$

11. (CBSE 2017) Equations Reducible to Simpler Forms

Question: Solve the equation $\cos(2x) + 3\cos x = 1$.

Explanation:

1. Use Double Angle Identity: Substitute $\cos(2x) = 2\cos^2x-1$

   $$2\cos^2x-1+3\cos x=1$$ $$\implies 2\cos^2x+3\cos x-2=0$$

2. Solve the Quadratic: Let $y=\cos x$. The equation $2y^2+3y-2=0$ factors as $$(2y-1)(y+2)=0$$ The solutions are $y=1/2$ and $y=-2$.

3. Find Valid Solutions: This means $\cos x=1/2$ or $\cos x=-2$. The second case is impossible.

4. Find General Solution: For $\cos x=1/2$, the principal value is $\pi/3$. The general solution is $$x=2n\pi \pm \pi/3$$

Answer: $$x=2n\pi \pm \pi/3$$

12. (CBSE 2020) Equations Reducible to Simpler Forms

Question: Find the general solution of the equation $\sin(2x) + \sin(4x) = 2\sin(3x)$.

Explanation:

1. Apply Sum-to-Product to LHS: Use $$\sin A + \sin B = 2\sin((A+B)/2)\cos((A-B)/2)$$ $$2\sin(3x)\cos(-x) = 2\sin(3x)$$ Since $\cos(-x)=\cos x$, this is $$2\sin(3x)\cos(x) = 2\sin(3x)$$ $$\implies 2\sin(3x)\cos(x) - 2\sin(3x) = 0$$ $$\implies 2\sin(3x)(\cos x - 1) = 0$$

2. Solve the two cases:

   • Case 1: $\sin(3x)=0 \implies 3x=n\pi \implies x=n\pi/3$.

   • Case 2: $\cos x - 1 = 0 \implies \cos x = 1 \implies x=2n\pi$.

3. Combine Solutions: The second set of solutions, $x=2n\pi$, is a subset of the first, $x=n\pi/3$ (when $n$ is a multiple of 6). The complete solution is the larger set.

Answer: $x=n\pi/3$.

13. (CBSE 2015) Equations Reducible to Simpler Forms

Question: Solve $\cos(4x) = \sin(2x)$.

Explanation:

1. Convert to a single function: Use the cofunction identity $\sin(2x) = \cos(\pi/2 - 2x)$. The equation becomes $$\cos(4x) = \cos(\pi/2 - 2x)$$
2. Apply General Solution for Cosine: For $\cos A = \cos B$, the solution is $A=2n\pi \pm B$. $$4x = 2n\pi \pm (\pi/2-2x)$$
3. Solve the two cases:
   • Case 1 (+): $$4x = 2n\pi + \pi/2 - 2x$$ $$\implies 6x = (4n+1)\pi/2$$ $$\implies x = (4n+1)\pi/12$$
   • Case 2 (-): $$4x = 2n\pi - (\pi/2 - 2x)$$ $$\implies 2x = (4n-1)\pi/2$$ $$\implies x = (4n-1)\pi/4$$

Answer: The solutions are $$x=(4n+1)\pi/12$$ and $$x=(4n-1)\pi/4$$

14. (CBSE 2019) Equations of the Form $a\sin x + b\cos x = c$

Question: Find the general solution for $\sin x + \cos x = 1$.

Explanation:

1. Normalize: Divide the equation by $\sqrt{a^2+b^2} = \sqrt{1^2+1^2} = \sqrt{2}$. $$\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}}$$
2. Convert to a Single Function: Recognize that $1/\sqrt{2} = \cos(\pi/4) = \sin(\pi/4)$. We can use either the sin or cos sum formula. Let's use $$\cos(A-B) = \cos A \cos B + \sin A \sin B$$ $$\cos x\cos(\pi/4) + \sin x\sin(\pi/4) = \cos(x-\pi/4) = 1/\sqrt{2}$$
3. Solve: The equation is $$\cos(x-\pi/4) = \cos(\pi/4)$$ The general solution is $$x-\pi/4 = 2n\pi \pm \pi/4$$
   • Case 1 (+): $$x-\pi/4 = 2n\pi + \pi/4$$ $$\implies x=2n\pi+\pi/2$$
   • Case 2 (-): $$x-\pi/4 = 2n\pi - \pi/4$$ $$\implies x=2n\pi$$

Answer: $x=2n\pi+\pi/2$ and $x=2n\pi$.

15. (CBSE 2018) Equations of the Form $a\sin x + b\cos x = c$

Question: Solve the equation $\sin x - \cos x = -1$.

Explanation:

1. Normalize: Divide by $\sqrt{1^2+(-1)^2} = \sqrt{2}$. $$\frac{1}{\sqrt{2}}\sin x - \frac{1}{\sqrt{2}}\cos x = -\frac{1}{\sqrt{2}}$$
2. Convert: This can be written as $$\sin x\cos(\pi/4) - \cos x\sin(\pi/4) = \sin(x-\pi/4)$$ So, $$\sin(x-\pi/4) = -1/\sqrt{2} = \sin(-\pi/4)$$
3. Solve: The general solution for $\sin A = \sin B$ is $A = n\pi + (-1)^n B$. $$x-\pi/4 = n\pi + (-1)^n(-\pi/4)$$

Answer: $$x = n\pi + \pi/4 - (-1)^n (\pi/4)$$

16. (CBSE 2021) Equations of the Form $a\sin x + b\cos x = c$

Question: Find the number of solutions of $\cos x + \sqrt{3}\sin x = 2$ in the interval $[0, 2\pi]$.

Explanation:

1. Normalize: Divide by $\sqrt{1^2+(\sqrt{3})^2} = 2$. This gives $$\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = 1$$
2. Convert: This can be written as $$\sin(\pi/6)\cos x + \cos(\pi/6)\sin x = \sin(x+\pi/6)$$ So, $$\sin(x+\pi/6)=1$$
3. Solve: The only way sine can be 1 is if the angle is $\pi/2$ plus any full rotation. So, $$x+\pi/6 = 2n\pi+\pi/2$$ $$x=2n\pi+\pi/2 - \pi/6 = 2n\pi+\pi/3$$
4. Find Solution in Interval: For $n=0$, we get $x=\pi/3$. All other integer values of $n$ give solutions outside $[0, 2\pi]$.

Answer: There is 1 solution.

17. (CBSE 2017) Equations of the Form $a\sin x + b\cos x = c$

Question: For what values of k does the equation $3\cos x + 4\sin x = k$ have a solution?

Explanation: An expression of the form $a\cos x + b\sin x$ can be converted to $R\cos(x-\alpha)$ where $R = \sqrt{a^2+b^2}$.

Here, $a=3, b=4$, so $R = \sqrt{3^2+4^2}=5$. The expression is equivalent to $5\cos(x-\alpha)$, for some angle $\alpha$.

The range of the cosine function is $[-1, 1]$. Therefore, the range of $5\cos(x-\alpha)$ is $[-5, 5]$.

For the equation to have a solution, the value of $k$ must be within this range.

Answer: The solution exists for $k \in [-5, 5]$.

18. (CBSE 2022) Equations of the Form $a\sin x + b\cos x = c$

Question: Solve $\tan x + \sec x = \sqrt{3}$.

Explanation:

1. Convert and Simplify: $$\frac{\sin x+1}{\cos x} = \sqrt{3}$$ $$\implies \sin x+1 = \sqrt{3}\cos x$$ This gives $$\sqrt{3}\cos x - \sin x = 1$$

2. Normalize: Divide by $\sqrt{(\sqrt{3})^2+(-1)^2} = 2$. We get $$\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x = 1/2$$

3. Convert: This is in the form $$\cos A \cos B - \sin A \sin B = \cos(A+B)$$ Let $B=\pi/6$. The equation is $$\cos(x+\pi/6)=1/2 = \cos(\pi/3)$$

4. Solve: The general solution is $$x+\pi/6=2n\pi\pm\pi/3$$

   • Case 1 (+): $x=2n\pi+\pi/3-\pi/6 = 2n\pi+\pi/6$.

   • Case 2 (-): $x=2n\pi-\pi/3-\pi/6 = 2n\pi-\pi/2$.

We must check for extraneous solutions where $\cos x=0$. The second case, $x=2n\pi-\pi/2$, gives odd multiples of $\pi/2$, so it is rejected.

Answer: The general solution is $$x = 2n\pi+\pi/6$$

19. (CBSE 2023) Miscellaneous Examples

Question: Find the sum of solutions of $\cos(2x)=\sin x$ in $[0,2\pi]$.

Explanation:

1. Form a Quadratic: Use $\cos(2x)=1-2\sin^2x$. The equation becomes $$1-2\sin^2x=\sin x$$ $$\implies 2\sin^2x+\sin x-1=0$$.

2. Solve for $\sin x$: Factoring gives $$(2\sin x-1)(\sin x+1)=0$$ So, $\sin x=1/2$ or $\sin x=-1$.

3. Find Solutions in $[0,2\pi]$:

   • For $\sin x=1/2$, the solutions are $x=\pi/6$ and $x=5\pi/6$.

   • For $\sin x=-1$, the solution is $x=3\pi/2$.

4. Calculate the Sum: $$Sum = \pi/6 + 5\pi/6 + 3\pi/2$$ $$= 6\pi/6 + 3\pi/2$$ $$= \pi+3\pi/2=5\pi/2$$

Answer: $$5\pi/2$$

20. (CBSE 2015) Miscellaneous Examples

Question: Solve $\sin^4x + \cos^4x = 1/2$.

Explanation:

1. Use Algebraic Identity: $$\sin^4x + \cos^4x = (\sin^2x+\cos^2x)^2 - 2\sin^2x\cos^2x$$ $$= 1 - 2\sin^2x\cos^2x$$

2. Simplify: $$1-2\sin^2x\cos^2x = 1/2$$ $$\implies 2\sin^2x\cos^2x=1/2$$

3. Use Double Angle Identity: Multiply by 2: $$4\sin^2x\cos^2x=1$$ $$\implies (2\sin x\cos x)^2=1$$ $$\implies \sin^2(2x)=1$$

4. Solve: This means $\sin(2x)=\pm 1$. This occurs when $2x$ is an odd multiple of $\pi/2$.

   So, $2x = (2n+1)\pi/2$.

Answer: $$x=(2n+1)\pi/4$$

21. (CBSE 2018) Miscellaneous Examples

Question: Find the number of solutions of $|\cos x| = 2[x]$ where [.] is the GIF.

Explanation:

1. Analyze Ranges: The range of the LHS, $|\cos x|$, is $[0,1]$. The RHS, $2[x]$, produces only even integer values (…, -2, 0, 2, …).

2. Find Common Value: The only value that is in both the range of the LHS and the set of possible values for the RHS is 0.

3. Solve the System: We must have both sides equal to 0 simultaneously.

   • $|\cos x|=0 \implies \cos x = 0 \implies x = (2n+1)\pi/2$.

   • $2[x]=0 \implies [x]=0$. This is true for all $x$ in the interval $0 \le x < 1$.

4. Find Intersection: We need a solution for $\cos x=0$ that lies in the interval $[0,1)$. The smallest positive solution for $\cos x=0$ is $x=\pi/2 \approx 1.57$, which is not in the interval. There is no intersection.

Answer: There are no solutions.

22. (CBSE 2019) Miscellaneous Examples

Question: Find the general solution of the equation $2\sin x\tan x + 2\sin x - \tan x - 1 = 0$.

Explanation:

1. Factor by Grouping: $$(2\sin x\tan x - \tan x) + (2\sin x - 1) = 0$$ $$\implies \tan x(2\sin x - 1) + 1(2\sin x - 1) = 0$$ $$\implies (2\sin x-1)(\tan x+1)=0$$
2. Solve the two cases:
   • Case 1: $$2\sin x-1=0$$ $$\implies \sin x=1/2$$ The general solution is $$x=n\pi+(-1)^n\pi/6$$
   • Case 2: $$\tan x+1=0$$ $$\implies \tan x=-1$$ The general solution is $$x=n\pi-\pi/4$$

Answer: $$x=n\pi+(-1)^n\pi/6$$ and $$x=n\pi-\pi/4$$

23. (CBSE 2020) Miscellaneous Examples

Question: Find the sum of the principal solutions of the equation $$\cot^2x + \frac{3}{\sin x} + 3 = 0$$

Explanation:

1. Convert to a single function: Use $\cot^2x = \csc^2x - 1$. $$(\csc^2x-1) + 3\csc x + 3 = 0$$ $$\implies \csc^2x+3\csc x+2=0$$

2. Solve the Quadratic: Let $y=\csc x$. $$y^2+3y+2=0$$ $$\implies (y+1)(y+2)=0$$ So, $y=-1$ or $y=-2$.

3. Find $\sin x$ values: This means $\csc x=-1 \implies \sin x=-1$,

   or $\csc x=-2 \implies \sin x=-1/2$.

4. Find Principal Solutions in $[0, 2\pi)$:

   • For $\sin x=-1$, the solution is $x=3\pi/2$.
   • For $\sin x=-1/2$, solutions are in Q3 and Q4. Reference angle is $\pi/6$.

   Solutions are $x=\pi+\pi/6 = 7\pi/6$ and $x=2\pi-\pi/6 = 11\pi/6$.

5. Calculate the Sum: $$Sum = 3\pi/2 + 7\pi/6 + 11\pi/6$$ $$= 9\pi/6+7\pi/6+11\pi/6$$ $$= 27\pi/6 = 9\pi/2$$

Answer: $9\pi/2$.

24. (CBSE 2022) Miscellaneous Examples

Question: Solve $\tan\theta + \tan(2\theta) + \tan\theta\tan(2\theta) = 1$.

Explanation:

1. Rearrange the Equation: The structure resembles the tangent addition formula. $$\tan\theta+\tan(2\theta) = 1-\tan\theta\tan(2\theta)$$ $$\implies \frac{\tan\theta+\tan(2\theta)}{1-\tan\theta\tan(2\theta)}=1$$

2. Apply Identity: The LHS is the formula for $$\tan(\theta+2\theta) = \tan(3\theta)$$ So, $\tan(3\theta)=1$.

3. Find General Solution: The general solution for $\tan A=1$ is $$A=n\pi+\pi/4$$

   So, $3\theta=n\pi+\pi/4$.

Answer: $$\theta = n\pi/3+\pi/12$$

25. (CBSE 2021) Miscellaneous Examples

Question: Find the number of solutions of the equation $\cos x + \cos y = 2$ in the interval $[0, 2\pi]$.

Explanation: The maximum value of the cosine function is 1. For the sum of two cosine terms to be 2, both must be at their maximum value simultaneously.

• $\cos x = 1 \implies x=0, 2\pi$ in the interval $[0, 2\pi]$.
• $\cos y = 1 \implies y=0, 2\pi$ in the interval $[0, 2\pi]$.

The solutions are ordered pairs (x,y). We can form pairs using these values:

(0,0), (0, 2$\pi$), (2$\pi$, 0), and (2$\pi$, 2$\pi$).

Answer: There are 4 solution pairs (x,y).

26. (CBSE Sample Paper) Miscellaneous Examples

Question: Solve $3\tan^2x - 1 = 0$.

Explanation:

1. Solve for $\tan^2x$: $\tan^2x=1/3$.

2. Find Principal Value: $$\tan^2x = (1/\sqrt{3})^2 = \tan^2(\pi/6)$$ So, $\alpha=\pi/6$.

3. Apply General Formula: The general solution for $\tan^2x=\tan^2\alpha$ is $$x = n\pi \pm \alpha$$

Answer: $$x = n\pi \pm \pi/6, n \in \mathbb{Z}$$

27. (CBSE 2017) Miscellaneous Examples

Question: If $\sqrt{3}\tan(2\theta) + \sqrt{3}\tan(3\theta) + \tan(2\theta)\tan(3\theta)=1$, find the general value of $\theta$.

Explanation:

1. Rearrange: $$\sqrt{3}(\tan(2\theta)+\tan(3\theta))=1-\tan(2\theta)\tan(3\theta)$$

2. Form Tangent Sum: $$\frac{\tan(2\theta)+\tan(3\theta)}{1-\tan(2\theta)\tan(3\theta)}=1/\sqrt{3}$$

3. Apply Identity: The LHS is $$\tan(2\theta+3\theta) = \tan(5\theta)$$ So, $\tan(5\theta)=1/\sqrt{3}$.

4. Solve: The principal value for $\tan A=1/\sqrt{3}$ is $A=\pi/6$.

   The general solution is $5\theta=n\pi+\pi/6$.

Answer: $$\theta=n\pi/5+\pi/30$$

28. (CBSE 2019) Miscellaneous Examples

Question: How many solutions does the equation $3\sin^2x-7\sin x+2=0$ have in the interval $[0, 5\pi]$?

Explanation:

1. Solve the Quadratic: Let $y=\sin x$. $$3y^2-7y+2=0$$ $$\implies (3y-1)(y-2)=0$$ Since $\sin x=2$ is not possible, we solve $\sin x=1/3$.

2. Count Solutions per Cycle: The value $1/3$ is positive and less than 1. The equation $\sin x = 1/3$ has two solutions in each interval of length $2\pi$: one in Q1 and one in Q2.

3. Count Solutions in $[0, 5\pi]$: We break the interval down:

   • $[0, 2\pi]$: Two solutions.

   • $[2\pi, 4\pi]$: Two solutions.

   • $[4\pi, 5\pi]$: This interval covers Q1 and Q2 of the next cycle. Since $\sin x$ is positive in both these quadrants, there are two more solutions (one near $4\pi$ and one near $5\pi$).

   Total solutions = $2+2+2=6$.

Answer: There are 6 solutions.

29. (CBSE 2021) Miscellaneous Examples

Question: Solve for x: $\cos x + \sin x = \cos(2x)+\sin(2x)$.

Explanation:

1. Rearrange: Group the cosine and sine terms: $\cos x - \cos(2x) = \sin(2x)-\sin x$.
2. Apply Sum-to-Product Formulas: $$LHS = -2\sin(3x/2)\sin(-x/2)$$ $$= 2\sin(3x/2)\sin(x/2)$$ $$RHS= 2\cos(3x/2)\sin(x/2)$$
3. Solve the Equation: $$2\sin(3x/2)\sin(x/2) = 2\cos(3x/2)\sin(x/2)$$ $$\sin(x/2) [ \sin(3x/2)-\cos(3x/2) ] = 0$$
4. Find Solutions:
   • Case 1: $\sin(x/2)=0 \implies x/2=n\pi \implies x=2n\pi$.
   • Case 2: $\sin(3x/2)=\cos(3x/2) \implies \tan(3x/2)=1$.

This gives $3x/2=n\pi+\pi/4 \implies x=2n\pi/3+\pi/6$.

Answer: $x=2n\pi$ and $x=2n\pi/3+\pi/6$.

30. (CBSE 2023) Miscellaneous Examples

Question: Find the number of solutions of $\cos x = |1+\sin x|$ in $[0, 3\pi]$.

Explanation:

1. Simplify the Absolute Value: In the interval $[0, 3\pi]$, the value of $\sin x$ is always in $[-1, 1]$. Therefore, the term $1+\sin x$ is always non-negative ($[0, 2]$). So, $|1+\sin x|=1+\sin x$.

2. Solve the Equation: The equation becomes $$\cos x = 1+\sin x$$ $$\implies \cos x - \sin x = 1$$

3. Normalize: Divide by $\sqrt{1^2+(-1)^2}=\sqrt{2}$. $$\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x = \frac{1}{\sqrt{2}}$$

4. Convert and Find General Solution: This is $$\cos(x+\pi/4) = 1/\sqrt{2} = \cos(\pi/4)$$ The general solution is $$x+\pi/4 = 2n\pi \pm \pi/4$$

   • Case 1 (+): $x = 2n\pi$.
   • Case 2 (-): $x = 2n\pi - \pi/2$.

5. Find Solutions in $[0, 3\pi]$:

   • For $x=2n\pi$: $n=0 \implies x=0$; $n=1 \implies x=2\pi$.
   • For $x=2n\pi - \pi/2$: $n=1 \implies x=3\pi/2$.

Answer: The solutions are $0, 2\pi, 3\pi/2$. There are 3 solutions.