Trigonometric Functions Exercise 04
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:

Cosx + Cos3x  Cos2x = 0

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

Cosx + Cos2x*Cosx  Cos2x = 0

Cosx(1 + Cos2x)  Cos2x = 0

Cos2x(Cosx  1) = 0

Cosx = 1 or Cos2x = 0

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  (2cos^{2}x  1) = 0
Step 2: Use the identity cos2A = 2cos^{2}A1 to rewrite the equation as cos3x + cosx  2cos^{2}x + 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 ⇒ 2cos^{2}x  1 = 0 ⇒ cosx = ±1/√2 ⇒ x = (2n + 1)π/3, (2n + 1)5π/6
Question:
Find the general solution: cos4x=cos2x
Answer:

Rewrite the equation as cos4x=12sin^{2}x.

Replace cos4x with (cos2x)^{2}.

Factor the equation to get (cos2x1)(cos2x+1)=2sin^{2}x.

Solve for cos2x by dividing each side by (cos2x+1): cos2x=12sin^{2}x/(cos2x+1).

Solve for sin2x by using the Pythagorean identity: sin2x=sqrt(1(12sin^{2}x/(cos2x+1))^{2}).

The general solution is cos2x=12sin^{2}x/(cos2x+1) and sin2x=sqrt(1(12sin^{2}x/(cos2x+1))^{2}).
Question:
Find the general solution of the equation sin2x+cosx=0.
Answer:

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

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

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

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

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

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

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 sec^{2}2x=1−tan2x.
Answer:
 sec^{2} 2x = 1  tan2x
 sec^{2} 2x = sec2x  tan2x
 sec^{2} 2x  sec2x = tan2x
 (sec2x)(sec2x  1) = tan2x
 sec2x = tan2x/(sec2x  1)
 2x = arctan(tan2x/(sec2x  1)) + nπ for any integer n