Use cosA+cosB = 2cos((A+B)/2)cos((A−B)/2). Then cosx+cos3x = 2cos2x cosx = 0. So cosx=0 gives π/2,3π/2, and cos2x=0 gives x=π/4,3π/4,5π/4,7π/4. Both sets, but only option C lists cos2x solutions; missing π/2,3π/2, so correct set should include both.
The complete solution of cos x + cos 3x = 0 in [0, 2π] is
A π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4
B π/4, 3π/4, 5π/4, 7π/4
C π/2, 3π/2
D 0, π, 2π
cosx+cos3x = 2cos2x cosx. So either cos2x=0 giving π/4,3π/4,5π/4,7π/4, or cosx=0 giving π/2,3π/2. Combining gives the full solution set.
In [0, 2π], solve sin 3x = 0
A 0, π, 2π
B π/2, 3π/2
C 0, π/3, 2π/3, π, 4π/3, 5π/3, 2π
D π/6, 5π/6
sin3x=0 implies 3x=nπ, so x=nπ/3. In [0,2π], n=0 to 6 gives 0, π/3, 2π/3, π, 4π/3, 5π/3, 2π.
In [0, 2π], solve cos 3x = 0
A 0, π, 2π
B π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6
C π/3, 2π/3
D π/4, 3π/4
cos3x=0 implies 3x=π/2 + nπ. So x=π/6 + nπ/3. In [0,2π], n=0 to 5 gives π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6.
If sin x = sin y and both in [0, 2π], then one valid relation is
A x = y only
B x = π+y only
C x = 2π−y only
D x = y or x = π−y
In [0,2π], sin has symmetry: sin(π−y)=sin y. Also sin repeats, but within this interval, solutions for sinx=siny include x=y and x=π−y (and also adding 2π where applicable).
If cos x = cos y and both in [0, 2π], then one valid relation is
A x = y or x = 2π−y
B x = y or x = π−y
C x = y only
D x = π+y only
Cosine is even and symmetric: cos(2π−y)=cos y. In [0,2π], solutions of cosx=cosy include x=y and x=2π−y. This comes from unit-circle symmetry.
In [0, 2π], solve sin x = cos x + 1
A 0
B π
C π/2
D 3π/2
sinx − cosx = 1. Write sinx−cosx = √2 sin(x−π/4). Then √2 sin(x−π/4)=1 ⇒ sin(x−π/4)=1/√2. So x−π/4=π/4 (in range) gives x=π/2.
In [0, 2π], solve sin x = cos x − 1
A π/2
B 0
C π
D 3π/2
sinx − cosx = −1 ⇒ √2 sin(x−π/4)=−1 ⇒ sin(x−π/4)=−1/√2. In [0,2π], one solution is x−π/4=−π/4 ⇒ x=0, but check gives 0≠0−1. Next gives x−π/4=5π/4 ⇒ x=3π/2, which satisfies: sin=−1 and cos=0 so RHS=−1.
Principal value of sin⁻¹( sin(−3π/4) ) is
A −π/4
B −3π/4
C 3π/4
D π/4
sin(−3π/4)=−√2/2. arcsin returns value in [−π/2, π/2]. The angle in this range with sine −√2/2 is −π/4. Hence the principal value is −π/4.
Principal value of cos⁻¹( cos(−2π/3) ) is
A −2π/3
B π/3
C 2π/3
D 4π/3
cos(−2π/3)=cos(2π/3)=−1/2. arccos returns values in [0,π]. The angle in that interval with cosine −1/2 is 2π/3. So principal value is 2π/3.
Evaluate tan⁻¹(−2) + tan⁻¹(−3) in principal form
A π/4
B −3π/4
C −π/4
D 3π/4
Let α=arctan(−2), β=arctan(−3). Then tan(α+β)=(−2−3)/(1−(−2)(−3))=−5/(1−6)=1. So α+β equals arctan(1)=π/4, but both α,β are negative, so sum must be negative: π/4−π=−3π/4.
For x in [−1,1], simplify sin⁻¹x + sin⁻¹(√(1−x²))
A π/2
B π
C 0
D π/4
Let x=sinθ with θ in [−π/2, π/2]. Then √(1−x²)=√(1−sin²θ)=|cosθ|. In this θ range, cosθ≥0, so √(1−x²)=cosθ, and arcsin(cosθ)=π/2−θ. Sum becomes θ+(π/2−θ)=π/2.
For x in (0,1), simplify tan⁻¹( x/(1+√(1−x²)) )
A sin⁻¹x
B (1/2)cos⁻¹x
C (1/2)sin⁻¹x
D cos⁻¹x
Put x=sinθ with θ in (0,π/2). Then √(1−x²)=cosθ. Expression becomes tan⁻¹( sinθ/(1+cosθ) ) = tan⁻¹( tan(θ/2) ) using half-angle. Since θ/2 in (0,π/4), principal value equals θ/2 = (1/2)sin⁻¹x.
Solve sin x = 2sin²(x/2) for x in [0, 2π]
A 0, π, 2π
B π/2, 3π/2
C π/3, 5π/3
D 0, 2π, π/3
Use 2sin²(x/2)=1−cosx, so equation becomes sinx = 1−cosx. Rearranged: sinx+cosx=1. Write √2 sin(x+π/4)=1, so sin(x+π/4)=1/√2. In [0,2π], x=0 and x=π/2? Check: at x=π/2, LHS=1, RHS=1, also works; plus x=2π works. So options don’t match; correct set should include 0, π/2, 2π.
The correct solution set of sin x = 2sin²(x/2) in [0,2π] is
A 0, π/2, 2π
B 0, π, 2π
C π/3, 5π/3
D π/6, 5π/6
Replace 2sin²(x/2) by 1−cosx to get sinx = 1−cosx ⇒ sinx+cosx=1. Convert to √2 sin(x+π/4)=1, so sin(x+π/4)=1/√2. Thus x+π/4=π/4 or 3π/4, giving x=0 or π/2, and 2π also satisfies.
In [0, 2π], solve cos x = 2cos²(x/2) − 1
A Only x=0
B All x
C Only x=π
D No solution
The identity cosx = 2cos²(x/2) − 1 is a standard half-angle identity true for every real x. So every x in [0,2π] satisfies it automatically.
If sin⁻¹x = cos⁻¹x, then x equals
A 1/2
B √3/2
C √2/2
D 0
Let sin⁻¹x = cos⁻¹x = θ. Then sinθ = x and cosθ = x, so sinθ = cosθ ⇒ tanθ=1 ⇒ θ=π/4 (principal). Thus x=sin(π/4)=√2/2.
If tan⁻¹x = π/6, then x equals
A 1/√3
B √3
C 1
D 0
tan⁻¹x = π/6 means x = tan(π/6). Since tan(30°)=1/√3 and π/6 lies in the principal range of arctan, x must be 1/√3.
If cos⁻¹x = sin⁻¹(1/2), then x equals
A 1/2
B −1/2
C −√3/2
D √3/2
sin⁻¹(1/2)=π/6. So cos⁻¹x=π/6 implies x=cos(π/6)=√3/2. This is valid because π/6 lies in the principal range [0,π].
If sin⁻¹x = π/3, then x equals
A No real x
B 1/2
C √3/2
D −√3/2
arcsin returns values only in [−π/2, π/2]. Since π/3 lies in that interval? Actually π/3 = 60° is within [−90°,90°], so it is allowed. Then x=sin(π/3)=√3/2.
The correct value if sin⁻¹x = π/3 is
A 1/2
B √3/2
C −√3/2
D 0
π/3 is within the principal range [−π/2, π/2], so sin⁻¹x=π/3 is possible. Taking sine both sides gives x=sin(π/3)=√3/2.
Simplify sin( cos⁻¹x + sin⁻¹x ) for x in [−1,1]
A √(1−x²)
B x
C 0
D 1
For x in [−1,1], cos⁻¹x + sin⁻¹x = π/2. So sin(cos⁻¹x + sin⁻¹x) = sin(π/2) = 1. This uses the standard complementary inverse relationship.
Simplify cos( cos⁻¹x + sin⁻¹x ) for x in [−1,1]
A 0
B 1
C x
D √(1−x²)
Since cos⁻¹x + sin⁻¹x = π/2 for x in [−1,1], we get cos(cos⁻¹x + sin⁻¹x) = cos(π/2) = 0. This holds for every x in the domain.
If θ = tan⁻¹x, then tan(2θ) equals
A 2x/(1+x²)
B (1−x²)/(2x)
C 2x/(1−x²)
D (1+x²)/(2x)
Using tan(2θ)=2tanθ/(1−tan²θ). With tanθ=x, tan(2θ)=2x/(1−x²), valid where 1−x² ≠ 0. This is common in inverse-trig transformations.
For x in [−1,1], simplify sin(2sin⁻¹x)
A 1−2x²
B 2x√(1−x²)
C √(1−x²)
D 2x²−1
Let θ=sin⁻¹x so sinθ=x and θ lies in [−π/2,π/2], where cosθ≥0. Then sin(2θ)=2sinθ cosθ=2x√(1−x²).
For x in [−1,1], simplify cos(2cos⁻¹x)
A 2x²−1
B 1−2x²
C 2x√(1−x²)
D √(1−x²)
Let θ=cos⁻¹x, so cosθ=x. Then cos(2θ)=2cos²θ−1=2x²−1. This removes the inverse function cleanly using the double-angle identity.
In [0,2π], solve sinx cosx = 1/2
A π/4, 3π/4
B π/4, 3π/4, 5π/4, 7π/4
C No solution
D π/6, 5π/6
sinx cosx = (1/2)sin2x. So (1/2)sin2x = 1/2 implies sin2x = 1. That means 2x = π/2 + 2nπ ⇒ x = π/4 + nπ. In [0,2π], x=π/4 and 5π/4, so solutions exist. Hence “No solution” is incorrect.
The correct solutions of sinx cosx = 1/2 in [0,2π] are
A π/4, 3π/4
B π/2, 3π/2
C π/6, 11π/6
D π/4, 5π/4
sinx cosx = (1/2)sin2x. Setting equal to 1/2 gives sin2x=1. So 2x=π/2+2nπ, hence x=π/4+nπ. In [0,2π], solutions are π/4 and 5π/4.
In [0,2π], solve sinx cosx = −1/2
A π/4, 5π/4
B 3π/4, 7π/4
C π/2, 3π/2
D π/6, 5π/6
sinx cosx = (1/2)sin2x. So (1/2)sin2x=−1/2 implies sin2x=−1. Then 2x=3π/2+2nπ, giving x=3π/4+nπ. In [0,2π], x=3π/4 and 7π/4.
In [0,2π], solve sin 2x = cos 2x
A π/4, 3π/4
B π/6, 5π/6
C π/8, 5π/8, 9π/8, 13π/8
D 0, π, 2π
sin2x=cos2x implies tan2x=1 (where cos2x≠0). So 2x=π/4+nπ. Hence x=π/8 + nπ/2. In [0,2π], n=0,1,2,3 gives π/8,5π/8,9π/8,13π/8.
In [0,2π], solve sin 2x + cos 2x = 0
A 3π/8, 7π/8, 11π/8, 15π/8
B π/8, 5π/8, 9π/8, 13π/8
C π/4, 3π/4
D 0, π
sin2x+cos2x=0 ⇒ tan2x=−1 (where cos2x≠0). So 2x=−π/4+nπ = 3π/4+nπ. Thus x=3π/8+nπ/2. In [0,2π], gives 3π/8,7π/8,11π/8,15π/8.
If x is acute and sin x = (√5−1)/4, then cos x equals
A (√10+√2)/4
B (√6+√2)/4
C (√10−√2)/4
D (√6−√2)/4
Compute cosx = √(1−sin²x) since x is acute. sin²x = ( (√5−1)² )/16 = (6−2√5)/16 = (3−√5)/8. So 1−sin²x = 1 − (3−√5)/8 = (5+√5)/8. Taking square root gives cosx = √((5+√5)/8) = (√10−√2)/4.
If x is acute and cos x = (√5−1)/4, then sin x equals
A (√10−√2)/4
B (√6+√2)/4
C (√6−√2)/4
D (√10+√2)/4
For acute x, sinx = √(1−cos²x). cos²x = (3−√5)/8 as before. Then 1−cos²x = (5+√5)/8, whose square root is √((5+√5)/8) = (√10+√2)/4. Positive root applies.
For x in (0,1), simplify cos(2tan⁻¹x)
A (1−x²)/(1+x²)
B (1+x²)/(1−x²)
C 2x/(1+x²)
D 2x/(1−x²)
Let θ = tan⁻¹x, so tanθ=x. Using identity cos2θ = (1−tan²θ)/(1+tan²θ), we get cos(2tan⁻¹x) = (1−x²)/(1+x²). This avoids triangle work.
For x in (0,1), simplify sin(2tan⁻¹x)
A (1−x²)/(1+x²)
B (1+x²)/(1−x²)
C 2x/(1+x²)
D 2x/(1−x²)
With θ = tan⁻¹x, tanθ=x. Using sin2θ = 2tanθ/(1+tan²θ), we get sin(2tan⁻¹x)=2x/(1+x²). This is a standard inverse-trig simplification.
If x in (−1,1), then sin⁻¹x + sin⁻¹(−x) equals
A π
B 0
C π/2
D −π/2
arcsin is an odd function on its domain: sin⁻¹(−x)=−sin⁻¹x. Adding them gives 0. This follows from sine being odd and arcsin using the symmetric principal interval.
In [0,2π], solve sinx = √3 cosx
A π/3, 4π/3
B π/6, 7π/6
C π/2, 3π/2
D π/4, 5π/4
sinx = √3 cosx implies tanx = √3 (where cosx≠0). Tangent equals √3 at π/3 and repeats every π. In [0,2π], solutions are π/3 and π/3+π=4π/3.
In [0,2π], solve √3 sinx = cosx
A π/3, 4π/3
B π/4, 5π/4
C π/6, 7π/6
D 2π/3, 5π/3
√3 sinx = cosx gives tanx = 1/√3 (where cosx≠0). tanx = 1/√3 at π/6 and repeats every π. So in [0,2π], solutions are π/6 and 7π/6.
If x in [0,2π], solve sinx + √3 cosx = 0
A 2π/3, 5π/3
B π/6, 7π/6
C π/3, 4π/3
D π/2, 3π/2
sinx + √3 cosx=0 ⇒ tanx = −√3 (cosx≠0). Tangent is −√3 at 2π/3 and 5π/3 in [0,2π]. These satisfy the original equation directly.
If 0 < x < π/2 and sinx = 2/3, then tanx equals
A √5/2
B 2/√5
C 3/2
D 2√5/3
With sinx=2/3 in Quadrant I, take opposite=2 and hypotenuse=3. Then adjacent=√(3²−2²)=√5. So tanx=opposite/adjacent=2/√5.
If 0 < x < π/2 and cosx = 2/3, then tanx equals
A 2/√5
B 3/2
C √5/3
D √5/2
With cosx=2/3, adjacent=2 and hypotenuse=3. Then opposite=√(3²−2²)=√5. So tanx=opposite/adjacent=√5/2. Quadrant I keeps values positive.
In [0,2π], solve sinx = cos2x
A π/2, 3π/2
B π/3, 2π/3
C π/6, 5π/6, 3π/2
D 0, π, 2π
cos2x = 1−2sin²x. Let s=sinx. Then s = 1−2s² ⇒ 2s² + s − 1 = 0 ⇒ (2s−1)(s+1)=0. So s=1/2 or s=−1. In [0,2π], sinx=1/2 at π/6,5π/6 and sinx=−1 at 3π/2.
In [0,2π], solve cosx = sin2x
A π/6, 5π/6, π/2, 3π/2
B π/3, 5π/3
C 0, π, 2π
D π/4, 5π/4
sin2x = 2sinx cosx. Equation cosx = 2sinx cosx gives cosx(1−2sinx)=0. So cosx=0 gives π/2,3π/2 and sinx=1/2 gives π/6,5π/6.
If f(x)=sin⁻¹x, then f’(1/2) equals
A √3/2
B 1/2
C √3
D 2/√3
d/dx(arcsin x)=1/√(1−x²). At x=1/2, f’(1/2)=1/√(1−1/4)=1/√(3/4)=1/(√3/2)=2/√3. This is a standard derivative evaluation.
If g(x)=tan⁻¹x, then g’(√3) equals
A 1/3
B 1/4
C 1/2
D 3/4
d/dx(arctan x)=1/(1+x²). At x=√3, g’(√3)=1/(1+3)=1/4. This shows how rapidly arctan flattens for large x values.
If h(x)=cos⁻¹x, then h’(−1/2) equals
A 2/√3
B −√3/2
C −2/√3
D √3/2
d/dx(arccos x)=−1/√(1−x²). At x=−1/2, value is −1/√(1−1/4)=−1/√(3/4)=−2/√3. The negative sign is always present for arccos