Question:

If sinx+sin3x+sin5x=0, then the solution is A x=2nπ​/3,nϵI B nπ​/3,nϵI C (3n±1)π​/3,nϵI D none of these

Answer:

Answer: C (3n±1)π​/3,nϵI

Question:

Solve: tanx=−√3

Answer:

Step 1: Solve for x in terms of the inverse tangent: x = tan^-1(-√3).

Step 2: Use a calculator to calculate the value of x, which is approximately -54.735610317245 degrees.

Question:

Principle solution of equation cotx=−√3 is A π​/3 B 2π​/3 C π​/6 D 5π/6

Answer:

Step 1: Write the equation in terms of sine and cosine.

cotx = -√3

=> sinx/cosx = -√3

Step 2: Solve for sinx.

sinx = -√3cosx

Step 3: Use the fact that sin2x + cos2x = 1 to solve for cosx.

cos2x = 1 + (√3)2cos2x

=> cos2x = 1 - 3cos2x

=> 4cos2x = 1

=> cos2x = 1/4

Step 4: Solve for x.

cosx = ±1/2

=> x = ±π/3 or ±2π/3

Therefore, the principle solution of the equation cotx=-√3 is A π/3.

Question:

Solve the following equation: cosx+cos3x−cos2x=0

Answer:

1. Cosx + Cos3x - Cos2x = 0

2. Cosx + Cos(2x + x) - Cos2x = 0

3. Cosx + Cos2x*Cosx - Cos2x = 0

4. Cosx(1 + Cos2x) - Cos2x = 0

5. Cos2x(Cosx - 1) = 0

6. Cosx = 1 or Cos2x = 0

7. x = 0 or x = ±π/2

Question:

Find the general solution: cos3x+cosx−cos2x=0

Answer:

Step 1: Use the identity cos(A + B) = cosAcosB - sinAsinB to rewrite the equation as cos3x + cosx - (2cos²x - 1) = 0

Step 2: Use the identity cos2A = 2cos²A-1 to rewrite the equation as cos3x + cosx - 2cos²x + 1 = 0

Step 3: Factor out the common factor of cosx to get cosx(cos2x + 1) = 0

Step 4: Set each factor equal to zero and solve for x: cosx = 0 ⇒ x = (2n + 1)π/2 cos2x + 1 = 0 ⇒ 2cos²x - 1 = 0 ⇒ cosx = ±1/√2 ⇒ x = (2n + 1)π/3, (2n + 1)5π/6

Question:

Find the general solution: cos4x=cos2x

Answer:

1. Rewrite the equation as cos4x=1-2sin²x.

2. Replace cos4x with (cos2x)².

3. Factor the equation to get (cos2x-1)(cos2x+1)=2sin²x.

4. Solve for cos2x by dividing each side by (cos2x+1): cos2x=1-2sin²x/(cos2x+1).

5. Solve for sin2x by using the Pythagorean identity: sin2x=sqrt(1-(1-2sin²x/(cos2x+1))²).

6. The general solution is cos2x=1-2sin²x/(cos2x+1) and sin2x=sqrt(1-(1-2sin²x/(cos2x+1))²).

Question:

Find the general solution of the equation sin2x+cosx=0.

Answer:

1. Start by expanding the left side of the equation: sin2x+cosx=2sinxcosx+cosx

2. Set the left side of the equation equal to zero: 2sinxcosx+cosx=0

3. Factor the left side of the equation: (2sinx+1)(cosx)=0

4. Set each factor equal to zero: 2sinx+1=0 and cosx=0

5. Solve for each factor: 2sinx=-1 and cosx=0

6. Find the solution for each equation: x= -π/4 and x=π/2

7. The general solution of the equation sin2x+cosx=0 is x=-π/4 and x=π/2.

Question:

Find the general solution of cosec x = -2.

Answer:

Step 1: cosec x = -2

Step 2: 1/sin x = -2

Step 3: sin x = -1/2

Step 4: x = arcsin(-1/2) + 2πn, where n is an integer.

Step 5: The general solution is x = arcsin(-1/2) + 2πn, where n is an integer.

Question:

Find the general solution of the equation sec²2x=1−tan2x.

Answer:

1. sec² 2x = 1 - tan2x
2. sec² 2x = sec2x - tan2x
3. sec² 2x - sec2x = -tan2x
4. (sec2x)(sec2x - 1) = -tan2x
5. sec2x = -tan2x/(sec2x - 1)
6. 2x = arctan(-tan2x/(sec2x - 1)) + nπ for any integer n